 Hello. In our last lecture, we had discussed some examples of which involved the use of Lorentz transformation. We specially discussed the problem of mumeuzon shower, which provided an experimental support to special theory of relativity. Classically, it was never understood that out of the mumeuzons, which are created at upper end of the atmosphere, how so many of them are able to reach earth, though their lifetime was so small, they should have decayed in between. It was special theory of relativity, which could explain this behavior on the basis of time dilation that provided a real good experimental support to special theory of relativity. So, this is what we have written. We discussed some examples involving use of Lorentz transformation. Then, we discussed the problem of mumeuzon shower and explained how time dilation helped us to understand it. Now, let us go ahead to a slightly new topic, which is the velocity transformation. The idea is as follows that you know the velocity of a particular particle in S frame and you want to find out the velocity of the same particle as being seen in S frame. So, let us be clear that there is one particular particle, which is moving somewhere and an observer S is trying to observe that particular particle. Similarly, an observer in S is also trying to observe the same particle. The particle is neither fixed in S nor fixed in S, it could be anywhere and in fact that particular particle need not move with a constant velocity. It could even be accelerated particle, it does not matter. What we are saying is that instantaneous velocity of this particular particle is being measured by both the observers. One observer in S, another observer in S and of course both S and S observers are in their respective frames of references which are inertia. So, this is what we call as velocity transformation. We know the components of the velocity of a particle in S and want to find the same in S for the same particle. So, the equations which will relate these velocity components in S to the velocity components in S would be called velocity transformation. Let us put some rotations because these are the things which one has to be very clear. See symbol v, we will reserve for the relative velocity between the frames because both the frames are inertial. Therefore, v has to be constant. We cannot change as a function of time otherwise at least one of the frame is non-inertial. So, v is constant as a function of time. Now, this observer S tries to observe the behavior of a particle. Let us write it here. So, this is an observer in S frame and there is some particular particle which is moving in some arbitrary direction and the observer in S frame also tries to observe the same particle. So, whatever is the velocity measured of this particular particle in S frame that will be called u. Whatever is the velocity measured of this particular particle in this particular frame will be called u prime while the relative velocity between these two frames will be called v. So, these are the symbols that we will be using in this particular lecture or in this particular series of lectures. Repeating here, v is the relative velocity between the frames which is constant as a function of time. u is the instantaneous velocity of the particle. Even if it is accelerated, we can always find out an instantaneous velocity. This need not be constant. While u prime is the instantaneous velocity of the same particle which is being now observed in S frame of reference and this need not be constant. Let us talk about the events related to displacement. We have generally agreed that in special theory of relativity, it is important to talk in terms of events. So, what we are saying, let us assume for the moment that the velocity is only along the x direction. We will generalize it later. So, a particle is moving along x direction in S frame and we have to find out its location at two different times. So, let us suppose this particular particle at time t 1 is found at a coordinate x 1. This I call as event 1. If this same particle is observed at a time t 2 at coordinate x 2, then this particular event of the particle being at x 2 at time t 2 is called event 2. So, these are my two events. Event 1 particle found at x 1 at t 1, event number 2 particle found at x 2 at time t is equal to t 2. So, these are my two events. Now, you would agree that if I have to define the instantaneous velocity of the particle, what I must do? I must take x 2 minus x 1 which is the difference in the x coordinates, divide by the time difference. If the velocity of the particle would have been constant, then x 2 minus x 1 divided by t 2 minus t 1 would have anyway given me the velocity of the particle. But, if the particle velocity is constantly changing, then this time limit should be as small as possible in order to find out the instantaneous velocity. So, what we say that we define delta x as x 2 minus x 1 delta t as t 2 minus t 1. And in the time limit that delta t tends to 0, this quantity delta x divided by delta t will tend to become the instantaneous velocity of the particle. So, this is what I have written. Even if the velocity of the particle is not constant, delta x divided by delta 2 defined as x 2 minus x 1 divided by t 2 minus t 1 in the limit delta t tending to 0 would give the instantaneous velocity of the particle. Now, let us assume that the particle is having general motion in three dimension. It could need not move only along the x direction, it could move in any other direction. So, similarly, exactly in the same way, I can look at the displacement along the x direction in a given time, displacement along the y direction in the same time, displacement along the z direction in the same time and divide these things to get the instantaneous velocity components of this particular particle. So, this is what I have defined in the next transparency. If the motion is in general in three dimensions, then u x will be defined as delta t, limit of delta t tending to 0 delta x divided by delta t. Similarly, u y will be defined as in the limit delta t tending to 0 delta y divided by delta t and u z will be in the limit delta t tending to 0 delta z by delta t. These are the standard classical definitions of the instantaneous velocities and I do not think that we should have any problem in understanding it. The only thing which you have to realize that if the instantaneous velocity of the same particle is to be measured in S frame of reference, then of course, delta x has to be measured in S frame of reference. But now, because the time is also relative, therefore, this time also has to be measured in S frame of reference. See like, the observer in S cannot calculate the instantaneous velocity by taking delta x, he has to take delta x. Similarly, the instantaneous velocity cannot be calculated in S frame of reference by using delta t. It has to be delta t prime, which should be used. A observer in a particular frame of reference has to be consistent. All the information must be used in his own frame, in his or her own frame of reference to evaluate things like velocity. Hence, we define exactly in a same fashion the velocity components of the particle as u x prime. I have put a prime because it means this is being measured in S frame of reference in the limit delta t prime tending to 0 delta x prime divided by delta t prime. Similarly, u y prime will be in the limit delta t prime tending to 0 delta y prime divided by delta t prime, u z prime in the limit delta t prime tending to 0 delta z prime divided by delta t prime. I repeat, I have been using here delta t prime. Here, I am using delta t prime and not delta t because all these observations are being measured in S frame of reference. Of course, we agree that if delta t tends to 0 delta t prime will also tend to 0. Now, what is my problem? My problem is to find a relationship between u x prime, u y prime, u z prime in terms of u x, u y, u z or the other way. If I know u x, u y, u z, I want to find out what is u x prime, what is u y prime, what is u z prime. This is what will be called the velocity transformation. For doing that, this let me just read it here what I have said. Note that like displacement, the time difference has also to be measured in one's own frame. This is what I have emphasized. Now, for doing the velocity transformation, I would first like to write Lorentz transformation in differential form. Though we have been using it, but we have not been talking about it in detail, so let us spend little bit of time to understand it. See, we had earlier written that x prime should be equal to gamma x minus v t. This is the first equation of the Lorentz transformation. Now, if there are two events, I can write this as x 1 prime, this I can write as x 1 and this I can write as v t 1. Similarly, there is a second event, we write this as x 2 prime is equal to gamma x 2 minus v t 2. Now, here of course, x 1, x 2, t 1, t 2 have to be measured with respect to a fixed origin and all these conditions we have defined earlier. Now, what I say, let us take the difference of these two. If I take the difference of these two, I will call it x 2 prime minus x 1 prime will be equal to gamma x 2 minus x 1 minus v t 2 minus t 1. So, this becomes in the difference in the value of x. So, I can write this as delta x should be equal to gamma. This I can write again as delta x. This should be delta x prime. This is delta x minus v delta t. This is what I call as writing the Lorentz transformation in the differential form. The advantage here is that this depends only on the differences. Therefore, the origin of x and origin of t need not be defined because that things get cancelled out. So, eventually we are just looking at the differences. Similarly, I can also find out using the time transformation equation, then equation relating to delta t prime and delta t. The basic idea is that put just the deltas in terms of all these variables in front of all these variables. So, this is what I have written in this particular transparency where I have used Lorentz transformation in differential form. So, delta x prime is gamma delta x minus v delta t thing which we just now discussed delta y prime is equal to delta y delta z prime is equal to delta z and delta t prime will be gamma delta t minus v delta x upon c square. So, this is what is called writing Lorentz transformation in differential form. Now, in order to calculate a relationship between u x prime and u x, what I should be doing is using this particular delta x prime use this delta t prime. I divide these two, take the limit delta t prime tending to 0. This will give me u x prime and let us try to work out with these equations so that I get a relationship of u x prime in terms of u x. Similarly, I will use delta y prime and divide by delta t prime then somehow manipulate these two equations in order to relate u y prime to u y. Finally, delta z prime divided by delta t prime which in the limit delta t prime tending to 0 will give me u z prime manipulate these two equations to get the relationship in terms of u z. So, let us first look at these two equations delta x prime and delta t prime. So, I just divide delta x prime by delta t prime delta x prime is gamma delta x minus v delta t delta t prime is gamma delta t minus v delta x by c square. So, next transparency what I will have is just this equation divided by this equation which is what I have written here. This is delta x prime divided by delta t prime is equal to you remember delta x prime was delta x minus v delta t delta t prime was delta t minus v delta x divided by c square. So, all I have done is just divided these two. What I will do now? I will divide the numerator and denominator of this equation by delta t. So, if I divide these two by delta t in the numerator first term is delta x. So, dividing by delta t this becomes delta x divided by delta t. Then we have the second term v delta t if we divide this by delta t it just remains becomes phi the delta t goes away. Similarly, denominator if I divide by delta t delta t divided by delta t will give me 1. So, there is a 1 here. Then we have v upon c square then delta x divided by delta t can be taken out separately. Now, both the sides I will take the limit delta t tending to 0 which also implies delta t prime tending to 0. In that case this particular equation will tend to u x prime as we have discussed earlier. This will tend to u x as we have discussed earlier and of course, this will also tend to u x. So, the equation of the velocity transformation for the x component becomes u x prime is equal to u x minus v. This is u x minus v divided by 1 minus v upon c square is anyway there. So, v into this quantity which is u x v u x by c square. So, I have obtained the first equation of the velocity transformation which is the x component of the velocity. Similarly, for finding out the y component transformation instead of delta x prime I have to use delta y prime. Now, you remember delta y prime was anyway equal to delta y, but delta t prime was not equal to delta t. Therefore, in principle u y prime will not be equal to u y and that is what we see in the next transparency. So, we have here delta y prime divided by delta t prime. Delta y prime was equal to delta y as we know. Delta t prime we have already written the equation earlier is same gamma delta t minus v delta x divided by c square. We do the same trick divide numerator and denominator by delta t. This numerator becomes delta y divided by delta t. The denominator gamma divided by delta t becomes 1 minus v c v by c square into delta x divided by delta t. So, u y prime if we take the limit delta t tending to 0 which implies delta t prime tending to 0. This equation this particular part tends to u y this tends to u x. And therefore, the whole equation u y prime becomes equal to u y divided by gamma 1 minus v u x upon c square. So, what we see that we do not get u y prime equal to u y as is expected because remember even classically u x prime did change when we change from s to s prime. But as far as y components where y and z components were concerned they did not change. Hence, it appeared that the particle speed of light was becoming different. Now, because remember these equations are going to be valid even for the speed of light. So, if u x has to change u y and u z also have to change in order to make the speed of light same. And that is what we are precisely seeing here. The other observation which I would like to measure that u y prime this particular equation not only depends on u y also depends on u x. So, this equation is a little more involved because it depends both on u y and u x. Now, finding the z component is exactly identical to the y component and that is what we have written in the next equation. All that has happened that is instead of y we have z and therefore, we get the z component of the velocity transformation as u z prime is equal to u z divided by gamma multiplied by 1 minus v u x divided by c square. So, this is where we have written all the equations involving velocity transformation u x prime is equal to u x minus v divided by 1 minus v u x by c square u y prime is equal to u y divided by gamma 1 minus v u x by c square u z prime is equal to u z divided by gamma 1 minus v u x divided by c square. Remember this quantity here is same quantity in the bracket here. Here in the denominator we do not get divided multiplied by y while here the denominator gets multiplied by y. So, what I just to remember u x prime is equal to u x minus y was the classical transformation what has happened in terms of after inclusion of Lorentz transformation that this quantity gets divided by 1 minus v u x by c square. u y prime is equal to u y was the classical transformation here what is happening that we are getting gamma this whole quantity is divided by gamma multiplied by same quantity 1 minus v u x by c square u z prime u z divided by gamma into same quantity 1 minus v u x by c square. So, this is what we call as the velocity transformation relativistic velocity transformation. We can always find out inverse transformation because many times we use inverse transformation it means the information about the velocity is given in s frame of reference and we have to find out the velocities in s frame of reference. Prescription is simple change prime to unprime change unprime to prime change v to minus v. So, this is what we get as inverse velocity transformation u x is equal to u x prime plus v divided by 1 plus v u x prime by c square u y is equal to u y prime divided by gamma multiplied by 1 minus v u x prime by c square u z is equal to u z prime divided by gamma multiplied by 1 plus v u x prime divided by c square. Remember there is a plus sign here there is a plus sign here there is a plus sign here because there is also a plus sign here because we have changed v to minus v. It can be shown so at the moment we will not show it we will show these these things little later and we will have little more discussion on this particular aspect, but one can show if we take the magnitude of the velocities that if the speed of the particle is found to be less than c in s it will also be found to be less than c in s irrespective of v. If u is equal to c in s you will turn out to be equal to c in s prime also irrespective of v this is thing which we have always been mentioning that the speed of light should be frame independent quantity that is the pretext on which we had started special theory of relativity. Now, let us take one simple example of what we call as a relative velocity. We will take one more example and we will see that you know the classically many times we are bit confused about the relative velocity concept. Here once in relativity when we see a relative velocity it means very clearly that we change from one frame to another frame. So, let us say my question is that that an observer in s frame observes two particle of course the speeds of the particle have to be relativistic if we have to apply relativistic relativity theory. This particular particle moves according to this observer s with a speed of 0.8 c and let us I am just calling this particular particle as particle a. Let us assume there is another particle b which is moving to the left hand side with a speed of 0.9 c. Both these speeds are being measured by an observer in s. So, once I make a statement that this particular particle is moving with a speed of 0.8 c and this particular particle is moving to the left with a speed of 0.9 c it essentially means that we are talking all those things with respect to some other observer which is this observer s. Now, if I have to ask the question what is the relative speed between particle a and particle b. It essentially means that I have to change my frame of reference. I go to let us say frame a, find out what is the speed that an observer sitting on a would find of particle b that is what is called relative speed. There is no other meaning of relative speed. Of course, if we wanted we could have also gone from s to b and found out what is the particle speed of a in b's frame. That will also be the relative velocity and as one can see very easily that this would have given me the same result other than the sign. So, relative velocities between these two particles will always be same as being observed by this particular either this particular observer or this particular observer. So, question now is that if these two speeds are given what is the speed of particle b in particle a's frame of reference which I am calling as s frame of reference. Once I calculate this quantity that by definition is the relative speed between a and b. Now, if I have to go from s to s frame of reference v which we have reserved the symbol v which has been reserved for the relative velocity between the frames. This v would be equal to the velocity of a as seen in s frame because I am changing from frame s to s, this s frame is linked to particle a. Therefore, v must be equal to 0.8 c. Of course, this is a vector. So, we can take an appropriate direction and put it vector if we want it to be more accurate. Now, this observer notices that this particular particle is moving to the left with a speed of 0.9 c and my question is that what is the speed of this particle as measured by an observer in s frame of reference. So, the particle whose speed is known in s and what I want to find out in s is this speed of 0.9 c and because this is in minus x direction I will write u x is equal to minus 0.9 c. Now, question is that I know the particle velocity which is 0.9 c with a negative sign and I know the velocity, relative velocity between the frames which is 0.8 c. What is u x prime? What is the velocity of this particular particle as being measured in s frame of reference? That is what will give me the answer of relative velocity. So, first equation on the top this particular transparency shows very clearly v is equal to 0.8 c as we have written here, u x is equal to minus 0.9 c that is what we had written earlier, u y is equal to 0, u z is equal to 0 because there is no velocity component in y or z direction the particle is assumed to be moving only in the x direction. We apply the formula of velocity transformation. Let me write the formula just to be sure u x prime is equal to u x minus v divided by 1 minus u x v upon c square. So, I will be substituting the values of u x and v in this particular formula to get value of u x prime. Here, I write u x prime is equal to u x which is minus 0.9 c then minus v is 0.8 c. So, there is minus sign here, there is a minus sign here. So, this minus sign I have taken out. So, this becomes 0.9 c plus 0.8 c divided by 1 minus u x once I say 1 minus u x, u x is minus 0.9 c. So, this becomes plus, this becomes 0.9 c into v which is 0.8 c. So, this is 0.8 c divided by c square. This c, this c get cancels with the c square, get cancel with the c square. So, you get 0.9 into 0.8 which is 0.72 to add 1, this becomes 1.72. So, my answer is if we add 0.9 and 0.8, we get minus 1.7 divided by 1.72 c which is equal to minus 0.988 c approximately. Of course, u i prime and u z prime are 0. So, we clearly see that the velocity components, the relative velocity turns out to be smaller than c. Even if I have changed the frame of reference, though classically we would have said that the relative velocity is 1.7, but here it does not turn out to be, it always turns out to be less than c as we had always expected. Let us go to another problem in which there are three particles and three frames involved and therefore, this problem is little more interesting from the simple problem which we have just now discussed. So, let us assume that we have a rod of proper length 1 meter, it is moving with a speed of 0.6 c in plus x direction as seen by an observer in a frame s. In the same frame, a particle is found to be moving in minus x direction that is the opposite direction to the rod with a speed of 0.8 c. So, this is the statement of the problem. We will see what are the things that are being asked and then we will show by a picture to make it clearer. So, what has been asked in the question? Find the time taken for the particle to cross the rod in a rods frame which I am calling as prime frame of reference, b in the frame s double prime of the particle. So, there is a third frame which is the particle frame and lastly in s frame which is the frame in which all these velocities have been given. So, there are three different frames, the ground frame let us call it, the s prime frame which is the rod frame s double prime frame which is the particle frame and essentially we have to find out in all these three frames what are the time taken or what is the time taken by the particle to cross the rod. We realize that the speeds that I am talking are all constant. So, we are in the we are actually discussing special theory of relativity. So, we can apply Lawrence transformation formulas or all the formulas derived on the basis of special theory of relativity without difficulty. So, this is what I have shown in this particular picture here approximately to give an idea that this is an observer s which I am calling as a ground observer. This picture has been shown with reference to this particular observer. This particular rod is moving with a speed of 0.6 c in the plus x direction. There is a particle which is moving in the left hand side with a speed of 0.8 c. This particular observer sitting here I am calling as s prime observer. This observer sitting here I am calling as s double prime observer. So, let us assume that this particular particle is slightly displaced with respect to this rod. So, it is not really hitting it. So, this particular particle can go over and cross this particular rod and my question is that how much time this particular particle would take to cross this particular rod? Let us define events. We will discuss about the observation in different frames of reference little later, but let us suppose this is my rod A B. So, I call my event number 1 is that when this particular particle reaches the B end of the rod, this I am calling as event 1. When this particular particle reaches here, then that is event B. Of course, how different observers will perceive this? That is we are going to discuss little later, but the events are being defined in this way that this particular particle reaches the end B and that is event number 1. When this particle reaches this end A of the rod, that is event number 2. In fact, these two events would signify because this is the point when this particular particle has just entered this particular rod area and this is the time when it is just leaving this particular rod area. So, in principle, the time difference between these two events should be the time that the particle takes to cross the rod. So, this is what I have written. Let me repeat here. Event number 1, particle reaching end B of the rod, event number 2, particle reaching end A of the rod. Can we apply the formula here? Yes, we can apply formula and this particular problem we will try to solve by applying time dilation and length contraction formula. But before that, we should answer the following question. Remember, whenever we are applying a formula either of time dilation or of length contraction, the first question that we must always ask which is the frame in which the information given, appropriate information given is proper. So, this is what I have explained here. Is there a frame in which time interval between these two events is proper? Is there a frame in which the length of the rod is proper? Answer to the second question is very simple because the length has to be proper in the rod's own frame because it is only in the rod's own frame that the length is stationary, the rod is stationary. So, length observed in the rod's own frame of reference will be a proper length. So, it is the S frame of reference in which the length of the rod is proper. And remember, what we have been given in the problem is the proper length. The length in other frames has not been given. So, if you want to find out, we will have to find it out by using an appropriate method, but it is only in S frame that length is proper. So, information given about the length is in S frame of reference. Now, let us try to answer the question about the time interval. Is there a time interval which is proper out of these three frames? Now, if you look in comparison to S frame, according to an observer in S frame, this particle is moving to the right, this rod is moving to the right and this particle is moving to the left. Both are moving. So, in principle, the event of this particular point, this particular particle crossing the rod and crossing the other rod will happen at two different positions. In S frame, also it will be different because according to an observer S frame, in fact, the situation between S and S frame is very similar to the plate form and the train example that we have discussed earlier. The person sitting on the rod will feel that the rod is stationary and the particle is actually going and coming first to the end A and then coming back, coming, going out of the end B. Therefore, these two events occur at different position even with respect to an observer in S. It is only in S double prime that is the particle frame in which that particular person will notice as if the rod is coming towards him and first the end A of the rod reaches to him and then end B of the rod reaches him to him and therefore both the observations are being measured or being found at the same place in S double prime. So, in principle, it is the time interval in S double prime frame of reference which is a proper time interval. This is what has been shown here in this particular transparency that it is this S frame of reference in which the length is proper. It is S double prime frame of reference in which time interval is proper. If you consider these two, if our problem was not mentioning about S frame of reference, these two would be just like the train and the compartment problem. The difference here is that we have not been given in a third frame in which this rod as well as this particular particle both are moving and we realize that in S frame neither the length nor the time interval is proper. Now, if we know the, if we have to find out the time interval, we should realize that we can go, we can use the time dilation formula only when one time interval is proper. So, it means it is possible if I know the time interval in S double prime, I can use the time dilation formula and find out time interval in S Similarly, if I know time interval in S double prime, I can also find out time interval in S frame of reference. If I know the relative velocity between S and S double prime, but I cannot directly go from S to S prime because none of these time intervals is proper. So, this is what I have shown in the next transparency. Let me first just read it. We can go from S double prime to S prime or from S double prime to S by use of time dilation formula, however we cannot go from S to S prime directly. So, this picture shows a flow of the use of time dilation formula. As we have said that time interval is proper in this particular frame of reference. So, I can go from here to here. By multiplying by gamma, I can find out what will be the time interval in this particular frame of reference. I can go from here to here backwards also, but then I have to divide by gamma. Similarly, I can go from here to here. If I know the time interval in this frame, I can also find out the time interval in this particular frame by multiplying by gamma. If I know the time interval in this frame, I can find out time interval in this frame by dividing by gamma. So, this particular path and this particular path is open, but I cannot directly go from S to S prime. So, I cannot multiply the time interval in S to gamma by gamma and go to S prime or take the time interval in S prime and multiply it by gamma to get into S frame because time interval is neither proper in this frame nor in this particular frame. Of course, if I have to go from S to S prime, I have to necessarily follow this particular route. I must go from here to this and then from here to here. Or if I have to go from here to here, I can go from this particular frame to this particular frame and then come back to here. I cannot make a direct application of the formula in this particular path because neither the time interval here is proper nor it is here which is proper. So, in order to apply time dilation formula, one of the time interval has to be proper. If we have understood this particular thing, then let us decide how do we go about this particular problem. Now, once we know the path, once we know the flow chart, then we can find out time interval in any of the frames and following the flow chart, we can always find out the time interval in any other frame. What I have written? We need to find out time interval in at least one of the frame. Then we can find in any other by use of formula being careful of the flow direction. That is very important that we have to be clear about the flow direction. Once we do that, then we can find out without any error. So, let us choose one of the frame. Let us try S prime frame of reference which is the rod frame of reference. So, let us first try to calculate the time interval that particle will take to cross the rod in S prime frame of reference. Then we will use this particular flow chart to find out the time interval in different frames. Now, let us look in this. What information has been given in S prime frame of reference? We have already agreed that in this particular frame, the length of the rod is proper because rod is stationary. Therefore, I know the length which is one meter which is given in S prime frame of reference. This has been given to me but I do not know any other speeds. In fact, I know the speed of S in S prime frame of reference but immediately that is not very useful to me because as I have said that I cannot directly go from S prime to S or from S to S prime. And in any case, I have to evaluate first in this particular frame the time taken by the particle to cross the rod. Therefore, what I need clearly is the speed of the particle as being measured in S prime frame of reference. So, if I have to calculate the speed of the particle, I come back to the problem that we have just now discussed earlier about relative velocity. I have to find out the relative velocity between S prime and S double prime. So, my problem is that if I know V that is the velocity of S prime observer in S frame, if I know Ux which is the velocity of the particle in S frame, what is the velocity of the particle in S prime frame of reference? So, I come back here because I want to transform from S to S prime. Remember, S was my ground observer. Therefore, V is the relative velocity between S and S prime frame of reference because these are the two frames between which the transformation is taking place. So, V returns out to be equal to 0.6 C. The particles whose speed is to be observed is going to be given by Ux and that is the particle which is this particular particle which is moving to the left in S even in S frame and that is minus 0.8 C which has been given. So, we apply the same formula Ux prime is equal to Ux minus V divided by 1 minus Ux multiplied by V divided by C square. Exactly the same formula and we will get this answer. We can be now probably quick. Let us just write this particular formula once more which was Ux prime is equal to Ux minus V divided by 1 minus Ux into V upon C square. I have taken Ux is equal to 0.6 C. V is equal to minus 0.8 C. Ux is equal to 0.6 C. V is equal to minus 0.8 C because of this minus sign this becomes plus because of this minus sign the numerator is minus and these two quantities get added up. So, my Ux prime becomes minus 1.4 divided by 1 plus which is the product of these numbers 0.48 and Ux prime will turn out to be 1.4 C divided by 1.48. Now, what an observer in S prime would note? According to him he does not see the motion. He sees the motion of this particular particle only he does not see his own motion. Therefore, situation according to him would be as if this particular particle is approaching towards him first it comes near B then goes and crosses over point A just like train crossing the platform. So, even B takes place here even A takes place here we have already agreed that in this frame time interval is not proper. But if I have to calculate the time how do I calculate the time? I know the speed of this particle in my frame remember I have to be consistent in my frame. I have found out the speed of this particle as minus 1.4 divided by 1.48 C I know the length in my own frame which is 1 meter which has already been given this length of course is a proper length. So, what I have to do is to just divide this one by this is speed to find out what is the time. So, this is the length 1 this is the speed Ux prime that will give me the time that this particular particle will take to cross the rod say as simple problem as in a traditional classical mechanics. In fact, in classical mechanics this will hardly be called a problem this is extremely simple problem. So, this is what I have written delta T prime is equal to L prime divided by Ux prime this L prime is 1 meter Ux prime we have already calculated is 1.4 C divided by 1.48. So, this goes to the numerator this becomes 1.48 divided by 1.4 C taking the value of C as 3 into 10 power 8 meters per second this delta T prime turns out to be 3.52 into 10 to the power minus 9 second. So, I have found out the time interval in S prime frame of reference. Now, if I have to find out S double prime I realize that this delta T prime is a dilated time. So, in order to go to S double prime I have to divide by gamma then I will find out its proper time interval which happens to be the time interval in S double prime frame. Once I know my proper time interval then I can always find out what is the time interval in S frame by following the flow. So, that is what I have written to find time interval in other frames first of all I must know the relative speeds and in fact, what we are going to use is the gamma. So, let us calculate the gamma values first. So, remember there is there are going to be 3 gamma values because one between S and S prime another between S prime and S double prime and third between S double prime and S. So, there are 3 gammas got their 3 relative velocities and therefore, we will calculate 3 different gamma values. So, this is the first equation here gamma S S prime it means I am finding out the value of gamma using the frames using the relative velocity between S and S prime frame of reference which is known as 0.6c and by now I think it should be comfortable that gamma S S prime will be given by 1 divided by under root 1 minus v square by c square while v will have to taking as 0.6c. So, this becomes 1 divided by under root 1 minus 0.6 square is 1.25. Similarly, gamma between S and S double prime S and S double prime frame have to be found out by using the relative velocity between S prime and S double prime between S and S double prime and that is 0.8c which has been given. So, this turns out to be equal to 5 by 3. Now, we find out the value of gamma between S prime and S double prime for which we had already calculated the relative velocity which was 1.48, 1.4c divided by 1.48c square cancels out. So, this gamma S prime S double prime turns out to be equal to 1.48 divided by 0.48 which is approximately equal to 3.08. So, I have found out corresponding to three different relative velocities, three different gamma values. Now, all we have to do is to apply the simple time dilation formula. I come back to the picture here. Relative velocity between these two was 0.6c, between these two was 0.8c, between these two was 1.4 divided by 1.48c that is what we calculated. Using this relative velocity, I got gamma which was this 1.48 divided by 0.48. Using this relative velocity, I got a value of gamma which was 1.25. Using this relative velocity which is 0.8c, I got gamma is equal to 5 by 3. But it is only in this frame that time interval is proper. We have agreed that I cannot use this particular root directly. So, I will not use this number directly. But I can always go from this frame to this frame. Remember, the flow direction is like this. It means time will actually be dilated here. This is proper time interval. So, if I have to find out along the reverse of the direction, I have to divide by gamma. If I have to go from this direction to this direction, I have to multiply by gamma. So, whatever is the time interval which has been found here, if I divide by gamma against the flow of the arrow, then I find out what is the time interval in this particular frame. Then in order to go from S double prime to S, I am following the arrow. So, I have to multiply by gamma in order to get the time interval in this particular frame of reference. So, I can go from here first in a direction opposite to the flow here in a direction towards the flow. I cannot use this channel directly. That is what is the idea. That is what I did. I calculated delta T double prime and used the delta T prime that I had just now calculated and divided by gamma S prime S double prime, because we are in a direction opposite to the direction of the flow of the arrow. This was delta T prime which was calculated earlier. Gamma S prime has been calculated here. If you calculate this time, this turns out to be 1.14 into 10 power minus 9 second. In order to go from S double prime to S frame to find out delta T in S frame, I have to multiply because I am in the same direction as the direction of the arrow. So, I have to multiply it by gamma S S double prime to get delta T double prime and this will be given by 5 by 3 into 0.48 divided by 1.4 C which turns out to be 1.9 into 10 power minus 9 second. I have found out all the three time intervals as we have calculated as we are asked. Now, there is another method of doing it. That is by finding out the length contraction formula. I do not want to directly find out or I do not want to directly use the time dilation formula, but I want to apply length contraction formula. Can I do that? Yes, I can do that. Also, there is another method of calculating all these time intervals by using the length contraction formula. So, let us try to do this particular thing here. Now, as far as length contraction is concerned, I realize that it is only this frame in which the time interval is proper. So, I am sorry, the length is proper. So, this is the frame in which length is proper. Now, I can go from this frame to this frame by applying length contraction formula. I can go from this frame to this frame by applying length contraction formula, but I cannot go from this frame to this frame because this channel is blocked because neither in this frame nor in this frame the length is proper. So, I must use proper length which is given in this frame and that has to follow this path or this path. I cannot use a path like this. Of course, I can use the path like this from here, go to here and come back here or go from here to here and come back here. That is possible, but not the direct use of length contraction formula from S to S double prime. That is prohibited because in neither of these two frames the length is proper. I already know the proper length which is 1 meter, which is an S frame of reference. I can immediately find out, I have already valued at the gamma values. So, problem is reasonably simple. I have to calculate the length in S double prime frame of reference in S frame of reference and both have to be obtained by dividing gamma because appropriate gamma because I am always in the direction of the arrow. Remember its length is contraction. So, the formula is dividing by gamma. So, L double prime will be L divided by gamma S double prime, which is as we have calculated gamma S prime S double prime was 1.48 divided by 0.48. So, this just becomes 0.48 divided by 1.48. Similarly, if I have to go from L prime to L, then I have to use gamma S S prime. This is 1.25. So, this becomes 1 divided by 1.25, which is 0.8 meters. So, I know the length in S frame as well as in S double prime frame. So, let us use this particular length first in S double prime frame of reference, which is comparatively simple. As far as S double prime frame is concerned, it means the frame of the particle. He would feel as if the rod is coming towards him, first end B reaches him, then end A reaches him and as we have discussed earlier, because both these events are occurring in the same position as far as S double prime is concerned. Therefore, this particular time interval turned out to be proper, but length is not proper in this particular case. So, delta T prime again is very simple. I have to know what is the length. I have to know what is the velocity with this rod is moving. I can immediately find out what will be the time interval. So, the length was 1 divided by 1.48 and V we had already calculated. This was the length and this was the V value. Immediately substitute, you get 1.14 into 10 power minus 9 second. Now, let us go in S frame where things are little more tricky. According to an observer in S frame, both particle and rod are moving. See in S prime or S double prime, there was one advantage that either particle or the rod was a crest, but in S frame, both particle and rods are moving. So, this particular particle is moving to the left side. This particular rod is moving to the right side. So, let us suppose event 1 occurs. It means this particular particle reaches this end of the rod. This is my event 1. Event 2 is when this particular particle reaches the second end of the rod, but between these two events, this particle is moving to the left, this rod is moving to the right. So, situation would be that this rod by the time the event 2 occurs, this rod would have already moved to the right and this particle was anyway moving towards the left. So, when event 2 occurs, it occurs somewhere here. Now, if you realize that in this particular frame, this let us suppose delta T is the time interval as observed between these two events as observed in S frame. So, during this particular time delta T, this rod would have moved to the right, a distance of 0.6C multiplied by delta T, while this particular particle would have moved to the left by an amount 0.8C into delta T. So, this is the distance that this particle has travelled in time delta T. This is the distance that this particular rod has travelled right in the time delta T. And as you can see very clearly from this picture, this length, this distance plus this distance must be equal to the length of the rod in case this event occurs at the first end of the rod and this event occurs at the second end of the rod. So, as far as S observer is concerned, I must get 0.8C multiplied by delta T plus 0.6C multiplied by delta T must be equal to the length of the rod as observed in S frame of reference. That is what I have written in the next transparency that 0.6C multiplied by delta T plus 0.8C multiplied by delta T is equal to the length which is equal to 0.8. We can add these things, you will get delta T is equal to 0.8 divided by 1.4C which really gets you 1.90 into 10 power minus 9 second, the result that we had obtained earlier from the time dilation method. Only one comment I would like to make that if you would have solved the classically the problem, probably would have said what is the relative speed and then we would have calculated classically the relative speed as 0.6 and 0.8 and divided by 0.8 divided by 1.4C. Now, would you say that relative velocity has increased and has become larger than C? No, that is not correct. Relative velocity in relativity means only one thing is that you must change your frame of reference and go to other frame of reference and then calculate the velocity. What I have calculated 1.4C is just a number because remember in relativity when I am saying in S frame I have not calculated the relative velocity, I have calculated the displacement of the particle and displacement of the rod and I have added them to make equal to the length of the rod as seen in S frame. I have not used the relative velocity, it just so happened that these two by adding gives me 1.4C, but it does not mean that this 1.4C can be interpreted as a relative velocity. For relative velocity I must change the frame. This is the point which I would like to see. So, this was essentially what we want to discuss in this particular lecture. So, at the end you would just like to give the summary. We obtained the velocity transformation and then eventually discussed some examples involving the velocity transformation. Thank you.