 Hello. So, let us recapitulate what we had discussed in our last lecture. We discussed two important consequences or I would say two most popular consequences of Lorentz transformation. One is length contraction, another is time dilation. We have defined something called a proper length that is the length of an object measured in a frame of reference in which the object is at rest. We call that particular length as the proper length. We said that if we go to any other frame of reference, the x component of this particular length will appear to be contracted in a different frame of reference. Of course, as we all know that the direction of x is always chosen along the relative velocity between the frames. The other aspect that we had discussed was time dilation. Similarly, we had defined quantity called a proper time interval which is a time interval between two events in a given frame in which the two events occur at the same position. This proper time interval appears to be dilated or supposed to be enhanced or increased if you go to any other frame of reference in which these two events do not appear at the same position. Then eventually we gave some examples relating to the length contraction. Today, we will try to give some more examples initially to discuss various other aspects of Lorentz transformation. Let us go to one example which is probably again a simple example. Let us assume that there is an observer standing on a platform, a train platform, the type of platform that we normally are quite aware of. Now, in that particular platform, a train passes by it. Let us assume that platform is also on a frame of reference which is inertial. Though we all know that earth is not strictly speaking a inertial frame of reference, but for all practical purposes we can treat it and let us assume that this particular platform is on an inertial frame of reference. And let us also assume that the train which is passing by the platform is also moving with constant velocity. Therefore, it also represents an inertial frame of reference. So, there is no acceleration involved. So, we can always treat one frame of reference as s and another as s prime. Now, both the observer, one observer on platform and another observer sitting in a compartment in a train, both of them want to measure the length of the platform assuming that this particular train is moving to a relativistic speed. It essentially means that the speed is very high close to speed of light. Then, let us have some discussion about the measurement of the length of the platform. So, this is what I have written as the subject of the example. Measurement of a length of a platform by an observer stationary on the platform and another, that is another observer moving in a train with relativistic speed relative to of course platform. Assume both frames to be inertial. So, this figure approximately represents the situation in this particular case. Let us assume that this is the platform. The beginning of the platform is A, the end of the platform is B and there is an observer standing on the platform. You can assume that this is the origin. You can assume it to be at one corner. We can do that thing and this represents the frame of reference S. This is a train which is moving with its velocity V, which is we are assuming to be relativistic speed, very high speed and an observer is standing or sitting in a particular seat in this particular compartment and let us call this frame of reference as S frame of reference. And as we said that both observer S and S prime want to measure the length of the platform AB. Now, let us define events. As we have always said that the relativity is much easier to solve problems if we define events and write their coordinates including time. So, that is what we will do. We will define two events relating to this measurement of length and similar thing which we have done also in the case of time dilation and then correction. So, we come to the fact when we define the events. My first event is defined when the origin of S prime coincides with the A end of the platform. It essentially means that whatever is the origin of S prime, we can take as one of the corner, let us suppose this particular corner coincides with point A. Of course, train is moving relative to point B. At a later time, the same point if we take for example, this as the origin of this particular S frame of reference, it makes so difference what we take. When this point coincides with point B of the platform, so that is my event number 2. So, I repeat event number A is that origin of S prime coincides with A, event number B at 2 is origin of S prime coincides with end B of the platform. So, these are my two events. Let us try to write the coordinates of these events in both S and S frame of reference. So, here I am defining my events, event 1, the origin of S coinciding with end A of platform. E 2, the origin of S coinciding with end B of the platform. We can choose any origin in S, it does not make a difference, because what matters is only the difference. But for thinking purposes, we can take either one end or somewhere in the center or at any other point. Let us write the events, but before that let me raise some questions. Is the coordinate difference equal to the length of the platform in any of the frames? This is question number 1 that I am posing. So, if we take the difference of x, let us assume that A B is along the x direction, because that is what we have to anyway assume, if you want to apply Lorentz transformation, because relative velocity direction is always chosen along the x direction or rather x direction is always chosen along the relative velocity direction. So, what is the coordinate difference that is x 2 minus x 1? Is it equal to, does it give equal to the length in any of the frame of reference S or S prime? The time difference between these two events, it means we take the time t 1 when event number 1 occurs, time t 2 when event number 2 occurs, we take the time difference t 2 minus t 1. Is it proper in any of the two frames that I have just now described namely S and S prime? Let us have a little bit of discussion. We realize that two events occur at different ends of the platform. One occur at N A, another occur at N B, this is what we have already always said and neither in S nor in S prime they occur at the same time. So, there is always going to be a time difference between event 1 and event 2, whether it is S which observes it or whether it is S prime observes it. And if you remember we have said it earlier that an object which is moving and if you measure the coordinates of the two ends, then that is not going to give you correct length. It is going to the coordinate difference is going to give you correct length only in a situation when the object is at rest because then only x 2 minus x 1 is correct length. Otherwise, if there is a time difference during the time difference of measurement the object has moved and the coordinate difference would not actually give the correct length. So, we realize that in this particular situation also it is only in the S frame that is the platform frame that it is only in the platform frame that the platform is not moving. In the train frame the platform is actually moving and I am measuring the length of the platform. So, if the coordinates of the two events has to give me the length of the platform, then this platform must be rest at rest because the two events are occurring at two different times and that is only the frame S. So, I conclude that in frame S the length of the platform will be given by the coordinate differences x 2 minus x 1. This is what I have written. Let us just read. The two events occur at the two different ends of the platform. However, time for two events is different both in S and S. We just now discussed. Hence, the coordinate difference will be equal to the length of the platform only in that frame in which platform is at rest. That is something which I have been probably over emphasizing it, but this is something which is to be understood because this is where people normally tend to make a mistake. They just write x 2 minus x 1 equal to length. x 2 minus x 1 is length only in a case the object is not moving or the time should be same. So, hence the coordinate difference will be equal to length of the platform only in that frame in which platform is at rest and that is the frame S. Now, if you go to S frame, suppose you are standing on a platform and you watch the train coming, you realize that event number 1 actually occurred at point A. Let us write here. Let us say this is A, this is B. So, suppose my observer was standing here. What he would observe? That train is coming from this side and passing by the platform and going to the other side. And let us suppose the front end is the origin of S. So, when this particular train reaches this particular object, this particular point, event 1 occurs. When the same train comes and reaches at point B, event number 2 occurs. So, position of event number 1 is A, position of event number 2 is B and as we have discussed, the position difference of this would be giving the length of the platform. But we realize that these two events actually occur at different location in S frame. It means that the time difference between them is not proper, because for the proper frame, for the time difference to be proper, these two events must occur at the same position. But what an observer S would be feeling? If you look from the point of view of an observer S, he would see a completely different situation. He would feel, because he does not see his motion. So, let us suppose this is my origin of the train. He will feel that this particular platform A, B is actually approaching towards him, because this person, observer sitting in S frame does not notice his or her own motion. So, according to him, the platform is actually coming and going by this side. This is a very common thing. If you are moving in a fast train and you see a particular station at which the train does not stop, it appears as if one end of the platform is coming towards you, then it passes by and the second end of the platform reaches you. So, in fact, this person standing here or sitting here will observe exactly the same thing that end A of the platform first reached him, then end B of the platform reached him. So, according to an observer in S frame, the first event that is A point meeting this particular point that occurs at this particular point, event number 2, that is point B meeting this particular point, again occurs at the same point. So, as far as the observer sitting on the train is concerned, he is sitting comfortably or standing comfortably at his place. He notices end of A of the platform coming towards him passing by, then end B of the platform coming towards him passing by. He is standing at his origin or he is sitting at his origin. For him, both the events point A coinciding with him, point B coinciding with him occurs at the same position. Hence, the time interval that will be measured by this observer in S frame of reference would really be a proper frame of reference, proper time interval. So, that is what I have written. In S, the two events occur at two different locations. What an observer in S would feel? An observer in S would feel, as I said, that they occur at the same position wherever the person is sitting or standing. Hence, time difference is proper in S frame of reference. So, therefore, in S frame, this time difference must be dilated. Now, let us go to frame S and try to write the coordinates. We just now agreed that if we are looking with the difference to the frame of reference S, which is the person standing on the platform, the two events, the difference of two events will give the actual length. Because according to this observer, this event occurs, let us say, at x 1, t 1. This event occurs at x 2, t 2. Now, clearly x 2, which is here, x 1, which is here, both these events, the position difference x 2 minus x 1, this minus this will actually give me the length of the platform. So, I must have x 2 minus x 1 is equal to l, which is the length of the platform. So, that is what I have written here. That in S, the length is proper and would be given by l is equal to x 2 minus x 1, where l I have taken as the length of the platform as measured in S frame of reference. However, this particular observer, S, had another way of measuring the length provided he knows the relative velocity of the train. If he knows with what speed the train is passing by it, then what he could have done, he could have just measured the time difference between the two events, even 1 and even 2. But for that, he must have the information of speed and he must have the information of the time differences. So, if he knows the time difference t 2 and t 1, t 2 minus t 1 rather, then if he multiplies by the speed of the train, he should still be able to get the length of the platform. Because that is another way of determining the length of the platform. If I know the speed of the train and I know how much time the train took, how much time the train took to go from here to here, that particular time and if I know the time difference between these two, then the speed multiplied by the time would give me the length of the platform. So, this is what I have written here that this length would also be given by V multiplied by t 2 minus t 1, which is the time difference between these two events. But as we have just now said that this time difference must be dilated because it is in the S frame of reference in which the time interval is proper. Therefore, this t 2 minus t 1 must be given by gamma times t 2 minus t 1 prime, where as we all know, gamma depends on the relative speed of the train and is actually given by gamma is equal to 1 upon under root 1 minus V square by c square. So, if I know the relative speed of the train, I also know the gamma. Once I know the gamma, I can substitute in this particular value gamma and I know that t 2 prime minus t 1 prime, that is the time difference measured in S frame of reference, which is actually the proper time interval between these two events that multiplied by gamma must give me t 2 minus t 1. So, this is equal to gamma times tau. If you remember, we have mentioned that normally we have been writing the proper time interval between two events as tau. So, this is gamma times tau. Now, let us go with respect to the frame of reference of S prime. We agreed that in S frame of reference, the time interval was really proper. But what about position difference? What about X 2 minus X 1? We have just now seen that as far as an observer in S prime is concerned, the two events occur at the same position. It means X 2 minus X 1 must be equal to 0. If that is the reason we have called that this time interval is a proper time interval because both the events occur at the same position. It essentially means that X 2 minus X 1 is 0, X 2 prime rather X 2 prime minus X 1 prime is 0. So, if we take the coordinate difference in S frame of reference, that coordinate difference is 0. So, again I remind you that X 2 prime minus X 1 prime is not equal to the length. So, because actually in the S frame of reference, its platform which is moving, therefore just taking the coordinate difference and putting equal to length can be disastrous. So, this is what I have written here. Time interval is proper in S frame of reference, tau is actually equal to T 2 prime minus T 1 prime and difference in X coordinate X 2 prime minus X 1 prime is equal to 0 which is definitely not equal to the length prime. It means the length measured, length of the platform measured in S frame of reference. Now, suppose you are sitting in a train and want to find out the length of the platform, what way for example, you can adopt. You can have a stopwatch in your hand and when N A comes, you can start this stopwatch. When N B arrives, you can stop the stopwatch, measure the time difference. Once you measure the time difference between these two events, that will be T 2 prime minus T 1 prime and as I have agreed, this is proper time interval. And if I know what is the speed of the platform relative to me in S which essentially is the same speed as an observer in S will notice of the train, if I know that particular v, then I can multiply the time difference between these two objects, these two events by velocity and I will get the length as observed in S frame of reference. So, this is what I have written here, the last line that L prime which is the length measured, length of the platform measured in S frame of reference will be given by v which is the relative velocity between the two frames multiplied by T 2 prime minus T 1 prime and as we had agreed that T 2 prime minus T 1 prime is tau which is the proper time interval between these two events. Therefore, this L prime can be written as v times tau which is the where tau is the proper time interval or time interval measured in this frame of reference itself. Let us just try to sort of compare our notes with reference to S and S frame of reference. We just now said that L prime must be given by v times tau, we have agreed that this tau is a proper time interval. Therefore, what an observer in S frame that is the platform frame would observe would be dilated time interval. Therefore, this tau must be given by T 2 minus T 1 divided by gamma where T 2 and T 1 are the times measured in the platform frame of reference, not the S frame of reference in the ground frame of reference where platform is at rest. So, this time interval divided by gamma must give me the proper time interval because T 2 minus T 1 was actually dilated tau is the shortest time interval. And if we agreed, we have said earlier that v multiplied by T 2 minus T 1 is actually the length measured in S frame of reference. So, I can write v T 2 minus T 1 as the length as measured in S frame. This turns out to be equal to L divided by gamma which tells that length of the platform as will be measured in a frame of reference S prime would turn out to be contracted. We have told that gamma is greater than 1. Therefore, L divided by gamma will give you a smaller length. So, this length will be contracted length as is expected because remember L is the proper length of the platform. Had it not been proper length, then you could not have applied this particular formula. When we apply length contraction formula, one of the length must be proper length. When we apply time dilation formula, one of the time interval must be proper time interval. Then only I can use that particular formula, otherwise I cannot use it. One that sees the contracted length in S frame as is expected. Now, let us discuss this another example. This example in some form or the other has been discussed in many textbooks. This is actually a real life example and before advent of special theory of relativity, this was considered as a problem which people were never able to understand. It is only special theory of relativity which provided a solution to this particular issue. Therefore, it has a historical importance. There are large number of particles, some of which are stable, some of which are unstable. Unstable means they decay and change into some other particles. Tom Mu-Mazon is one such particle. The lifetime of this particular particle is 2 microsecond if this particular particle is kept stationary or at rest in a particular frame of reference. What essentially the lifetime means, it means there is a statistical process of decay. It is like radioactive decay. When there is a particular chance, you can never predict where a particular nucleus will go under radioactive decay. It may take some time, it may take a different time. But you can always define something which is called the lifetime which represents in some way and approximately you can call it an average type of behavior. Essentially, the idea is that if lifetime is very large, that particle is much more stable. If the lifetime is very small, in comparison to that, the particle is likely to decay comparatively very soon. This particular particle has a very, very small lifetime which is 2 microseconds, 2 into 10 to the power of minus 6 seconds which is extremely small. Now, these particles appears are created at the upper atmosphere of the earth. Now, when they are created, they are created with a very large speed. The speed is essentially close to speed of light which in this particular problem we have taken to be approximately 0.998 C just to give some number approximately. So, it is a very large speed essentially close to speed of light. Now, we can calculate classically how much time it will take for this particular particle to reach earth. Then, we can approximately estimate that out of how many of these particular particles which are created at the upper atmosphere, how many of them would reach earth and that number turns out to be extremely small as we will just now see. But people have found that on earth, you get a very large number of new mesons or what we call it mumeoson shower from the upper atmosphere. People could not understand that if their lifetime was only 2 microseconds, we expected very small number of these mumeosons to actually have reached earth before that they would have decayed into something else. So, how it is possible that such large number of particles can really come and reach earth in spite of the fact that their lifetime is so small. And this particular problem was actually solved by special theory of relativity by virtue of time dilation. So, let us now read the problem. The incoming primary cosmic rays create mumeoson in the upper atmosphere. The lifetime of mumeosons at rest is 2 microseconds. If it is at rest, if we create this particular mumeoson in laboratory at rest, the lifetime will turn out to be 2 microseconds. If the mean speed of mumeoson is 0.998 C, let us talk of everything in terms of an average. What fraction of mumeoson created at the height of 20 kilometers? Let us take that, you know, they have created a distance of 20 kilometers above the mean sea level. How many of them would really reach the sea level? So, that is essentially the problem. Now, let us assume that we focus our attention on one particular mumeoson because we are looking at the average behavior and we will take one particular mumeoson which has exactly the same lifetime as 2 mumeosons, 2 microseconds. Now, if the particular particle is at rest, then we expect that after 2 microseconds, it will decay. So, if this particular mumeoson was created at the upper atmosphere of the earth and if there was a person sitting on mumeoson or my laboratory was moving along with the mumeoson, assuming of course, its speed to be constant because we are applying special theory of relativity, let us assume that mumeoson has the same speed as travelling through the earth, it remains constant as 0.998C. So, if it is always the same, then this particular observer would find that this particular mumeoson would decay in 2 microseconds. It is just like the train problem and the platform problem. Of course, my platform is from here 20 kilometers long from here to the top of the atmosphere. Now, this person who is coming along with the mumeoson would conclude that after 2 microseconds, this particular particle will decay and because the mumeoson is at rest in that particular frame of reference, so the event number 1 creation of mumeoson, event number 2 decay of mumeoson, they will occur at the same position and therefore, this time interval will be a proper time interval. However, on the earth, event number 1 occurs 20 kilometers above, which is the upper end of the atmosphere and the event number 2 occurs whenever it decays somewhere much down, much below down. So, according to the observer on earth, this time difference between these two events must be dilated and because the speed of the particle is very close to C, therefore, you would find that the time dilation effect will be very large and therefore, effectively an observer on earth would feel that the lifetime of the mumeoson has increased and that is why he could observe a such large number of fraction of mumeosons reaching earth. Now, let us do the numerical calculation. Decay time interval 2 into 10 to the power minus 6 second, which is 2 microsecond, is proper in mumeoson frame. In earth frame, it would appear to be dilated. If we take C is equal to 0.998 C and use this expression of gamma that we have just now written, gamma is equal to 1 upon under root 1 minus v square by C square. You put this number of v as 0.998 C, you will get gamma is equal to 15.82 approximately 16 of that order. It is a very large gamma, 15.82. Therefore, decay time or you can say lifetime of mumeoson in earth frame of reference will be just this vector multiplied by 2 into 10 to the power minus 6 second, which turns out to be approximately 3 into 10 to the power minus 5 second, one order of magnitude larger, 15 times larger. It makes lot of difference. Now, if you want to talk of the probability of decay, this essentially follows the same law or same rule as a nuclear decay or OBS, which is an exponential behavior, which is given here at the bottom end of this particular transparency that if n naught is the number of particles with which you had started, then total number of particles which remain at a time t is given by n divided by n naught to e raise to the power minus t by tau. Now, if we calculate, this is the fraction of the particles which will be reaching earth. Now, if we take this particular lifetime, we can calculate how much time it will take in s frame of reference, which I am calling as earth frame of reference, how much time it will take for this particular particle to reach earth, that will be 20 kilometers, which is here divided by what is the speed of the particle. So, normally if this particular mumeoson was supposed to reach earth, according to an observer on earth, that observer must have traveled this particular mumeoson must have traveled for time 6.68 into 10 to the power minus 5 seconds, approximately 7 into 10 to the power minus 5 second. So, if I have to find out what is the fraction, remember this calculation I am doing in earth's frame of reference. If I have to find out what is the fraction which must reach earth, that will be n upon n naught will be given by e raise to the power minus t by tau, where for t I will substitute 6.68 into 10 to the power minus 5 second and for tau we will substitute the value which we have obtained in my last transparency, which is 3.164 into 10 to the power minus 5 second. So, I substitute these numbers and next transparency I have given, this fraction turns out to be 0.12, which means approximately 12 percent of the mumeosons will be able to reach earth, others would have decayed in between, because this is sort of a statistical process in which some will decay, some will decay later, some will decay earlier. But if we do the, if we had done the classical calculation, it means for this particular time instead of, for this particular tau instead of using 1, sorry 3.164 into 10 to the power minus 5, you would have used 2 into 10 to the power minus 6, then this number would have turned out to be 3.12 into 10 to the power minus 15. So, remember there is an exponential factor, exponential factor is a very fast decaying function and you can see that how much the fraction has changed earlier. This was the calculation which people were doing earlier before special degree of relativity and they would have said that 1 into 10 to the power 15 approximately 1 out of 10 to the power 15 mumeosons would reach earth approximately. I am just ignoring the factor 3 just to calculate order of magnitude. That many number would reach earth while according to relativity 12 percent 12 out of 100 would reach earth. The number has become so large and that is what was experimentally observed which people could not understand earlier and only after special theory of relativity came people could understand it. That was considered as one of the great success of special theory of relativity. Just a question, what would an observer in a mumeoson frame will conclude? How you would identify this event in terms of these two events in terms of a mumeoson frame of reference as you can probably guess from that platform example which you have given just now that according to the observer on mumeoson the distance between the earth and the upper atmosphere will appear to be reduced or contracted. So, that observer will feel that this mumeoson has not actually travelled 20 kilometer distance but has travelled much smaller amount of distance that is why it has not decayed. Now, let us go back to our old example which we have discussed number of times in different ways that is the light emission from the center of train compartment. Remember first time when we had discussed we had discussed that instead of light we are throwing two balls one person sitting at the center of a train compartment throws two ball one in a direction of the motion of the train towards the front wall another towards the back wall and we defined two events when the ball reaches the front wall and the ball reaches the back wall. We concluded that in the frame of reference of train which we called as S prime the two events were stationary. Then we did a classical Galilean transformation and we found out that in S frame which is the ground frame of reference also these two events appear to be simultaneous. Second time when we discussed this particular problem was when we assumed that instead of the ball somebody is shining light one towards the front of the front wall another towards the back wall. We concluded at that time that these two events in the frame of reference of S prime which is the compartment frame of reference that the light reaching the front wall light reaching the back wall occurs at the same time and are simultaneous. Then at that time we discussed with reference to the ground frame and said that if second postulate of special theory of relativity was correct it means if the speed of light is same in S and S prime frame of reference then the observer in S frame of reference which is the ground frame of reference would feel that these two events are not simultaneous. But at that time we did not calculate time difference because we did not have Lorentz transformation. Now that we have our Lorentz transformation let us actually try to calculate the time difference between these two events which we have discussed just now as seen in the ground frame of reference which I am calling as S frame of reference. So let us look at this picture again. This is the picture of this particular train compartment. This person standing at the center of the platform he shines the light. Let us assume these two hands are just for making picture clearer. I have put them with a large hand long hands but let us assume that these are just at the same point and the light is emitted in this particular direction. This I am calling as a front wall light is being emitted towards back direction opposite direction to the motion of the train and this I am calling as a back wall. My event number one was light pulse reaching here, event number two light pulse reaching here which adds we had agreed in S frame of reference these two events are simultaneous. And now my question is that what will be the time interval between these two events as will be observed by an observer S standing on ground. We had earlier defined events. Let us redefine them. Let us rewrite them. Event number one light reaching the front wall of the train. Event number two light reaching the back wall of the train. So these are my events which we had already defined. Let us again reiterate them. Now what I will do because I want to do some mathematics using Lorentz transformation. Let us be clear and write coordinates of this event one and event two and because the initial information has been given to me in S frame of reference which is the frame of reference of the train. In fact, everything has been described with respect to that observer. So let us first write the coordinates in S frame of reference which is the train frame of reference. This is what I am doing in my next transparency. So, because this is S frame of reference all my coordinates are putting prime. Now that we know that lengths are different in two frames. So, this length which is being measured the length of the compartment itself that is in S frame of reference. So, that length also I am calling as L. So, we agree that if this is my train compartment and observer is just sitting standing in half the way then in this particular frame of reference this length is L. So, the first coordinate and this particular person is standing at the origin. So, we assume we get that this particular event even number one will occur at a distance of L prime by 2 positive side. I am taking this as positive side. This is negative side and this will occur at minus L prime by 2. So, this is what I have written here that x1 prime occurs at plus L prime by 2, x2 prime occurs at minus L prime by 2. T1 prime, time of the event number one is the time taken for the light to travel from here to here. We know that this time this light travels at the speed of light C. So, this particular time will be given by actually the length is L prime by 2 the distance it has to travel is L prime by 2. So, the time taken will be L prime by 2 C. So, time will be L prime by 2 C. Similarly, this light which has been emitted backwards also moves a distance of L prime by 2 and travels with the same speed C. So, therefore, this time interval will also be L prime by 2 C. So, according to an observer in S prime frame of reference this event number one occurs at a value of x L prime by 2 and a time L prime by 2 C. Event number two occurs at a coordinate of minus L prime by 2 and a time L prime by 2 C. These two times are same. These two events, therefore, are simultaneous in S prime frame of reference. This is what I have written here. In this particular transparency, x1 prime is equal to plus L prime by 2, T1 prime is equal to L prime divided by 2 C, x2 prime is equal to minus L prime by 2, T2 prime is equal to L prime by 2 C. Remember, we have not used any transformation to write these equations. I do not need any transformation if we have all the information in the same frame and all these basic information has been given in S prime frame of reference. I do not need any transformation. Let us try to write the coordinates of these two events in S frame. Let me first write the event number one. Event number one was light reaching the front wall. This occurred at L prime by 2 and L prime by 2 C. And if you remember, the Lorentz transformation gives you x prime is equal to gamma x minus VT. Now, the information has been given in x prime frame of reference and I want to prime in S frame of reference. Therefore, I must use inverse Lorentz transformation, which is given by x is equal to gamma x plus VT prime x prime plus VT. So, we said the prescription is change unprimed to prime, unprime to prime and prime to unprime change the sign from minus V to plus V, change the sign of V. So, this is my inverse Lorentz transformation. So, I use the inverse Lorentz transformation here to find out x 1 is given by gamma L prime by 2 plus V L prime by 2 C. This is my value of x prime. This is the value of my t prime L prime by 2 C. So, what we have written is x prime plus VT. This is what it comes to. So, I take gamma L prime by 2 out of this. So, this becomes gamma L prime by 2 multiplied by 1 plus V by C. Now, similarly we write inverse transformation for time. Inverse transformation for time will be given by T is equal to gamma T prime plus V x prime by C square. Again, using inverse transformation, we go back here to the transparency. T1 is equal to gamma times L prime by 2 C. This was the time plus because this is the inverse transformation. V relative speed between the frames x which is L prime by 2 divided by C square. I take gamma L prime by 2 C out of this bracket. This becomes 1 plus V by C. Of course, we have assumed that the origins are coincident when the light was omitted. So, at that instant of time, the observer which is standing on the ground was also at its origin and was just passing by. Now, let us write the coordinate of even number 2 in S frame of reference. You will realize that everything will be same here, exactly similar except that the L prime by 2 will change sign. So, this appears here and this appears here. This will change sign. So, that is what I am writing in the next transparency. Even number 2 in S frame. So, x2, that is the x coordinate of the second event will occur at gamma minus L prime by 2. That is where it observed. That was the x value plus V L prime by 2 C. Remember, time is same. Though event occurred at different location, but time was same. So, this L prime by 2 C. So, this remains the same thing. I take gamma L prime by 2 common and you get minus V plus V by C. We take T2, which is gamma T prime, which is same as L prime by 2 C minus V into x prime, which is minus L prime by 2 because this event occurred at the back of the train. So, this is minus L prime by 2 divided by C square. I take gamma L prime by 2 C i out. I get in bracket 1 minus V by C. So, as you can see that this time is different from the other time. In earlier case for the event 1, there was a positive sign here and this time interval was positive. This was 1 plus V by C. Here, it is 1 minus V by C. So, these two intervals are not simultaneous in S frame of reference, which is the ground frame of reference. As we had discussed earlier, but now we know what is the time difference. Actually, I can calculate T2 minus T1 and find out how much difference in the time was observed as according to S observer on the ground observer between these two events. So, that is what I have written in this particular transparency that we see that T2 is less than T1 because there is a negative sign here 1 minus V by C. So, T2 is less than T1. Hence, in S event 2 occurred before event 1. This also we had discussed earlier somewhat qualitatively without Lorentz transformation. We had said that if speed of light should remain same, then it should happen that because the light which travels backward has to travel a smaller distance. So, it will hit the wall first. The light which is emitted in front has to travel a larger distance, but with the same speed. Therefore, that event will occur later. So, that particular time will be larger. So, this is what we had qualitatively discussed. Now, we know their numerical numbers. So, hence in S event 2 occurred before event 1, as it was qualitatively discussed earlier, the time difference we can just calculate because that one factor will cancel out. So, that one of only the V by C thing has to be taken and this will turn out to be gamma L prime V by C square. So, this is the time interval between the two events as observed in S frame of reference. While in S frame of reference, this time interval was 0. Both these events occur at the same time. I have another question. Is the coordinate difference that we have calculated x2 minus x1 will be equal to the length of the compartment in S? I think we should be able to give quick answer that no, it cannot be. The reason is that these two events 2 occurred at two different ends of the compartment, but that compartment was moving and unless these two events occurred in S at the same time, x2 minus x1 will not give me correct length. x2 prime and x1 prime difference will give me the length in S frame of reference. But remember, in that particular frame of reference, any way the compartment was at rest. So, even the time difference, in the present example, in S frame of reference, these two events occurred at the same time. But in that frame of reference, the compartment was anyway at rest. Even if they would not have occurred at the same time, still x2 prime minus x1 prime would have given the correct length. But not in S frame of reference. In S frame of reference, these two events occurred at different time. Therefore, x2 prime minus x1 prime is not the correct length. Now, can we guess whether we are getting an over estimate or an under estimate of the length? Let us look into the future. Anyway, let us first, I have calculated the difference in the coordinates of the two events. This delta x will be given by x minus x2 because the event one was on the front of the wall. So, this turns out to be gamma L prime divided by 2 into 2. If we take the x coordinate transformation, which we have done just now, the result will turn out to be gamma L prime. So, according to the S observer, these two events occur at a spacing of gamma L prime along the x direction. The x1 minus x2 is an over estimate of length of the train because t1 is greater than t2 and can be collected. Let me just show you this particular picture. Let us suppose this was the position of the train when event number 2 occurred, which was an earlier event. So, this was x2. Even number 1 occurred when the light reached here. This event occurred at a later time. How much was the time difference was delta t, which we have just now calculated. Now, if I take the x1 event according to ground observer, this occurred here because the train has been moved and this train has moved by a distance of v delta t during the time interval in which the two events occurred according to S frame of reference. According to an observer in S frame, this event occurred earlier, this event occurred later and during this particular time, this train was actually moving. And how much distance it would have moved is the speed of the train multiplied by the time difference. So, this was the additional distance that this particular train moved. And therefore, x1 minus x2 actually gave me gamma L prime, which we have just now obtained. And if I want to determine the correct length, what I must do? From this gamma L prime, I must subtract v delta t. Then I will get the correct length as measured in S frame. This is what I have done in the next transparency. My L is equal to delta x minus v delta t. As we have just now seen, delta x is v gamma L prime, v is here. Delta t is gamma L prime v upon c square. We just redistribute the numbers. It becomes gamma L prime to bracket 1 minus v square by c square. If you know, gamma is equal to 1 upon under root v square minus c square and there is no under root here. So, this whole quantity will become L prime under root 1 minus v square by c square, which is 1 by gamma. So, I get length as L prime by gamma. So, as I expected, this length turns out to be a contracted length. This is what I had expected, that because in S prime frame of reference, the length was proper. Hence, it is in this particular frame of reference that the length will turn out to be contracted. Now, let us go to the conclusion. In this particular lecture, we have essentially discussed some examples of Lorentz transformation. In the next lecture, we will obtain the velocity transformation.