 Hello. In our last lecture, we had described time like and space like events. We had discussed that what are the basis on which it was postulated that speeds greater than speed of light would not be allowed in special theory of relativity. So, let us quickly recapitulate. We had given an example of two events and we had said that if delta x is the coordinate difference between these two events and delta t is the time difference between these two events assuming delta x and delta t to be positive. Then if delta x turns out to be smaller than c delta t, we call these events as time like separated events. Just to remember, when we say time like, it is time factor delta t which is large. Similarly, if delta x turns out to be larger than c delta t, then we call these events as space like separated events. When we say space like delta x is supposed to be larger, just to remember. And we had mentioned that in case we want the causality not to be violated, it means if in a frame of reference, two events occur. The second event is an outcome of the first event. First event is the cause of the second event. Then we would like that in any other frame of reference, we should always have the same order of the events. We cannot see in any other frame that whatever is the result or whatever is the outcome has come first and the reason for that is coming later. We have given an example that suppose I shoot somebody, then in my frame of reference, shooting of the gun if it is event number one and that person being hit by the bullet is event number two. Obviously, event number two has to occur after event number one. Then we should not be able to find out a frame of reference in which it so happens that that person observes that the person is hit by the bullet first and the bullet is shot later. So, the same order of events must be maintained if first event is the cause of the second event. So, this is what we call causality. So, we do not expect from the physics point of view that causality should be violated and that is possible only if we do not allow speeds greater than the speed of light. So, that is what is the physics reason for this particular conclusion. Just to mention again, I am writing here for time like separated events, time interval is pure and time order cannot be reversed. Of course, if it means that time order cannot be reversed, it means that we cannot find out a frame of reference in which delta t is 0. It means we cannot find a frame of reference in which the two events occur at the same time. Similarly, for space like separated events, space interval is pure and position order cannot be reversed. By again the same logic, if it has to be reversed, it means that we can also not find out the frame of reference in which these events occur at the same place. So, that is what I have written for space like separated events, space interval is pure and position order cannot be reversed. One cannot thus find a frame in which the two events occur at the same place. So, this is what we had discussed essentially in our last lecture. The most general behavior of these space like and time like, we will be discussing little later. So, let us postpone this particular discussion until we evolve certain other concepts. At the moment, what I would like to give is a few examples of space like and time like events and the cause and outcome related events just to make our ideas clear. So, let us go to one or two very, very simple examples. Let me take the example number one. Let us assume that an event occurs in a frame of reference S at origin at time 0. Any event it could be, for example, lightning striking or anything whatever we can imagine. So, one event occurs in a frame S at origin. What origin means x is equal to 0, y is equal to 0, z is equal to 0 and I also assume that this occurred at time t is equal to 0. Again, in the same frame S, a second event is found to be observed which is at a point A and the time difference, it means it occurs at 0.01 microsecond later. Micro is 10 to the power minus 6 second. So, it is 1 into 10 to the power minus 8 seconds later. Another event is found to be observed at point A not at the origin. And x coordinate of A has been given as 5 meters while y and z coordinates are same. So, we have two events, one occurring at x is equal to 0, another occurring at x is equal to 5. The first event occurring at time t is equal to 0 and the second event occurring at time t is equal to 0.01 microsecond. The question is, will it be possible to find a frame in which these two events would occur at the same place slash same time, find the speed of such a frame? So, first we have to resolve whether it should be possible to find a frame of reference in which these two events occur at the same place or at same time or whatever it is. And once we find out, then we have to find out the speed of that particular frame of reference in which these events either occur at the same place or at same time depending upon the situation. The problem is essentially simple. All we have to do is to take delta x and compare it with c delta t. If delta x happens to be larger than c delta t, then we know that is a space like event. If it happens to be smaller than c delta t, then we know that is a time like separated events. Then accordingly we can conclude whether it is possible to find a frame of reference in which these two events occur at the same position or at same time. So, we have been given everything. So, let us look at this particular thing. In S, the x and t coordinates of the events I have just listed. I have removed y and z, reference of y and z coordinates because that is not really material as far as this particular problem is concerned because in both the events y and z are 0. So, we take event number 1 which occurs at x is equal to 0 and t is equal to 0 as has been given in the question. We have event number 2 which occurs at point A which is at a distance of 5 meters along the x direction from the origin. So, x coordinate is 5 meters and as we have mentioned that it occurs at a time 0.01 microsecond, it means the time was 1 into 10 to the power minus 8 second. So, we have written my event table, my event number 1 is occurring at 0, 0 and event number 2 is occurring at 5 meters and 1 into 10 to the power minus 8 second. Now, all I have to do is to find out this delta x, then we have to find out this delta t, then multiply by c and compare the 2 that is all I have to do. Anyway, it is very easy to see that for the first event x was equal to 0, for the second event x was 5 meters. So, obviously delta x is 5 meters, x 2 minus x 1 is 5 meters. Similarly, t 2 minus t 1, t 1 occurred at time t is equal to 0, I mean event 1 occurred at time t is equal to 0, event number 2 occurred at time 0.01 microsecond. So, t 2 minus t 1 will be 1 into 10 to the power minus 8. If I multiply this by c to compare with this, and if I take c is equal to 3 into 10 to the power 8 will be 3, 10 to the power 8 will cancel to 10 to the power minus 8. So, this will become 3 meters. As I see very clearly that delta x is 5 meters, while c delta t is 3 meters. That is what I have written in this particular transparency. Delta x is equal to 5 meters, delta t is equal to 1 into 10 to the power minus 8 second, c delta t is therefore, 3 meters. Therefore, delta x is greater than c delta t. It means these events are separated space like, because delta x is turning out to be larger than c delta t. And if the events are separated space like, it is not possible to reverse the position order of the events. It means it is not possible to find a frame of reference in which these two events will occur at the same position. Though it is possible to find out the frame of reference in which these two events will occur at the same time. Let us just see and verify whatever we are telling. This is what I have mentioned just now. The events are space like separated. Hence, sign of space interval cannot be reversed, but that of time can reversed. How do we find this particular thing? Just use Lorentz transformation. This is what I have mentioned again. It is possible to find a frame in which the two events occur at the same time, but we cannot find one in which the events occur at the same place. Now, we go to Lorentz transformation. Assume that there is a frame of reference S, which is moving a relative to S with a velocity v. And I know that it is possible to have delta t is equal to 0. All I have to do is to find the speed of that particular frame of reference for which delta t will turn out to be 0. So, I write delta t is equal to gamma delta t minus v delta x by c square. This is a standard Lorentz transformation. Now, what I have to do? I know my delta t. I know my delta x. I have to find out a v for which delta t prime is equal to 0. So, I substitute delta t prime is equal to 0. Substitute there for delta t 1 into 10 to the power minus 8 second. I substitute for delta x 5 meters. Solve for v. This is what I have done in this particular transparency. I have written delta t prime is equal to gamma delta t minus v delta x by c square. For delta t, I have substituted 1 into 10 to the power minus 8 second. For delta x, I have substituted 5. For c square, I have substituted 9 into 10 to the power 16. Assuming of course, that c is equal to 3 into 10 to the power 8 meters per second. Just solve this equation. We get v is equal to 1.8 into 10 to the power 8 meters per second. Obviously, this speed is smaller than the speed of light. It means it is physically possible to find out the frame of reference in which these two events would occur at the same time. It means the time difference between these two events will turn out to be 0 in this frame of reference and that frame of reference would be moving relative to s with a speed of 1.8 into 10 to the power 8 meters per second. Obviously, it is clear that if we would have been larger than this particular value, if you look at this particular equation, if we would have been larger than this particular value, delta t prime would turn out to be negative and time order of the events would get reversed, which is possible in this particular case because the events are space like separated and not time like separated. Now, some of you can ask question, why cannot I do the same exercise and try to see whether I can really make delta x prime equal to 0 and if whatever we have said is correct, I should be able to get a speed greater than the speed of light, then only it will be possible for me to make delta x prime is equal to 0. So, let us just look into that to satisfy ourselves that whatever we are saying is correct. So, I do exactly the same thing instead of delta t prime making equal to 0, I try to make delta x prime equal to 0. So, let us try to find v, a speed of a frame of reference in which I assume, if at all possible, that these two events occur at the same position. So, now instead of writing delta, the transformation Lorentz transformation corresponding to time, I will write a Lorentz transformation corresponding to space. So, that is what it gives here, delta x prime is equal to gamma delta x minus v delta t, which is the Lorentz transformation corresponding to the x coordinate. I substitute my numbers, delta x was equal to 5 meters, I substituted here 5 meters, delta t was equal to 1 into 10 to the power minus 8 second, I substituted the value of 1 into 10 to the power minus 8 second here multiplied by v. I solve for this particular thing, you can see that gamma will turn out to be equal to 0. This 1, when divided by 5 will make this 5 and this 10 to the power minus 8, when it goes to numerator, it will become 10 to the power 8 and v will turn out to be 5 into 10 to the power 8 meters per second, which is obviously greater than speed of light. So, it means, if at all it was possible for us to have a frame of reference with v greater than 5 c, which in this particular case is 5 into 10 to the power 8 meters per second, then it was possible for us to find these two events at the same position. But as we have already seen that on the ground of physics and also to some extent on the mathematical ground, we have rejected speeds greater than the speed of light. Therefore, I would generally conclude that it is not possible to find out a frame of reference in which these two events occur at the same position. This is what I have written here. Hence, we see that it is not possible to find a frame with v less than c in which the events occur at the same place. Now, we have just now said that it is possible in this particular case to reverse the time order. It is possible to find out a frame of reference with v less than c in which the time order delta t turns out to be negative. It means event number 2 occurs before event number 1. We have said also that this is not possible if these two events are cause and outcome related. It means event number 1 and event number 2 have to be totally independent events. Event number 2 cannot be a cause of event number 1 if this particular condition has to happen. Let us just look and try to clarify our ideas little further. That is what I have written here. Can the two events described in the event be cause and effect type? For example, let us imagine in a particular case that this was my origin O and this was my point A and this distance is 5 meters. Everything being described in S frame. Now, let us assume that event number 1 was one car which is moving to the right was being found at origin. We have solved many such problems. So, let us assume that event number 1 was that this particular car was found at x is equal to 0 and this car is moving with a constant velocity v along the plus x direction. Let us assume that event number 2 is that this same car being found here at x is equal to A. Now, obviously these two events are cause and outcome related. If the car was not starting from this particular point, car would have never reached this particular point A. Therefore, for event number 2 to occur, event number 1 was essential. The car must have passed through this particular origin and must have come to point A. So, these two events are related. Event number 1 is sort of a cause for event number 2 because the car must pass through the origin then only it has to reach at A. Now, if this happens, can these two events lead to a situation which is given in this particular problem involving delta x is equal to 5 meters and delta t involving equal to 1 into 10 to the power minus 8 second. Is it possible that these two events are separated? Of course, delta x is given as 5 meters and the time interval between these two events was 1 into 10 to the power minus 8 second. That is what I have written here in this transparency. Can the two events described in the example be cause effect time? For example, can event 1 be a car passing by origin and event 2 be the same car passing at A? If this happens, it means and if the time separation between these two events is 100 to the power minus 8 second, it means the car must have taken 1 into 10 to the power minus 8 second to reach from O to A and the distance between O and A is 5 meters. Therefore, the speed of the car must be 5 meters divided by the time which is 1 into 10 to the power minus 8 second which will give you 5 into 10 to the power 8 meter per second, which is obviously larger than speed of light. Therefore, if these two events for example, were related like this, then this car must travel with speed greater than speed of light and if this is not happening, it is not possible in a situation like this for this particular situation to arise where delta x turns out to be larger than C delta t. And remember, only in a situation when delta x is larger than C delta t, it is possible to revert time interval. And because these two situations cannot lead to this particular delta x being greater than C delta t, therefore, we always say that time order cannot be reversed as far as these two events are concerned because these two events relating to the motion of the car cannot lead to this situation. And only if they would have led to a situation of delta x being greater than C delta t, it would have been possible for me to find their frame of reference in which the car would have reached A first before starting from O. Hence, you can see that by limiting the speed to a value less than or equal to C, we have avoided a situation in which causality would be violated. Same thing I have written here in this particular transparency. If that happens, the car has to travel a distance of 5 meter in 110 power minus 8 second as seen by an observer in S, which would mean its speed would be 5 into 10 power 8 meters per second, which is greater than C. Now, let us take a situation, make it a little more realistic. Let us assume that car is not really travelling with speed greater than speed of light. So, I change my delta x is fixed as 5 meters, I change my delta t. Let me still call my event number 1 as the car passing through O and event number 2 as car passing through point A, but I assign a speed to the car, which is less than speed of light. This is what I have written here. If car has to have speed less than C, then delta x has to be less than C delta t and the events would become time like separated. If the events are becoming time like separated, as we have discussed, it is not possible to revert the time order in any other frame of reference and therefore, it is not possible to find the violation of causality. It means we will not be able to find a frame in which the car reaches A first before starting from O. So, let us assume that the speed of car as seen in S is 2 into 10 to the power 8 meter per second. This is some value, which I have taken, some arbitrary value, which is of course less than speed of light. Of course, delta x is still 5 meters and if the car's speed happens to be 2 meters per second, 2 into 10 to the power 8 meters per second, I can find out how much time this particular car would take to reach point A, which is of course given by the distance, which is 5 meters divided by the speed, which is 2 into 10 to the power 8 meters per second, as I have described earlier. So, this time delta t would be given by this, which I can simplify, 5 divided by 2 is 2.5 into 10 to the power minus 8 second. So, in this case, we will find that delta t turns out to be equal to 2.5 into 10 to the power minus 8 second. Now, let us compare delta x with C delta t. So, what I have written here, time to reach A in S would then be given by 5 into 2 into 5 divided by 2 into 10 to the power minus 8, which is 2.5 into 10 to the power minus 8 second. I have to just multiply it by C to find out what is my C delta t. If I take C is equal to 3 into 10 to the power 8 meters per second, this 10 to the power 8 and this 10 to the power minus 8, their product will lead me 1. So, 2.5 multiplied by 3, which gives me 7.5 meters. Obviously, C delta t is larger than delta x and therefore, these are time like separated events. Now, it is possible to find out a frame of reference, in which these two events occur at the same position. But now, I will not be able to find a frame of reference in which these two events occur at the same time. Let us just see. The speed of the frame in which the two events occur at the same place. Now, I do exactly the same thing, but with slightly different numbers. Delta x is still 5 meters, but delta t has now become 2.5 into 10 to the power minus 8 second. I use exactly the same expression delta x is equal to gamma delta x minus v delta t delta x being 5 meters delta t being 2.5 into 10 to the power minus 8 second multiplied by v. I can solve for v. This gamma would of course, cancel, because there is a 0 here. This 5 divided by 2.5 will give me 2, which is 2 into 10 to the power 8 meters per second, which is of course, smaller than C. This result was also obvious. This turns out to be same as the speed of the car, which we know, because in the speed of the car, when if a person sitting in the car will observe both the events to be occurring at the same position. We have worked out many such problems earlier that if a person is sitting in the car, he will find that O is approaching him and the point A is approaching him. And as far as he is concerned, he or she is concerned, both the events are occurring at the same position. Therefore, delta x is 0, which I am able to find out. Of course, in this particular case, with a realistic speed, which is less than C, which is equal to 2 into 10 to the power 8 meters per second. But now, if I want to make delta t prime equal to 0, I will not be able to do it unless I allow speeds greater than the speed of light. Let us try and see. Of course, just the same narration, which I have said earlier. This is the same as the speed of the car as expected. Now, the speed of the frame in which the events could occur at the same time would be given as follows, which is given by delta t prime is equal to gamma delta t minus v delta x by C square. I substitute the new numbers. For delta t, I have 2.5 into 10 to the power minus 8 seconds. Delta x is same 5 meters. I work out this particular equation. I will get 4.5 into 10 to the power 8 meters per second, which is obviously larger than speed of light. So, only if there would have been a frame travelling with this much speed, then only it would have been possible for us to find out that the 2 events occur at the same time. Because of physical reasons, I have avoided, I have said the speeds like this are not allowed. Therefore, I would conclude that it is not possible to find a frame of reference in which the 2 events occur at the same time. And of course, it would mean that it will not be possible to find a frame of reference in which the time orders are reversed. Therefore, causality is accepted. It is not rejected. Causality has to be maintained. And in order to maintain this causality, as we have seen, that we have to restrict speeds to speeds less than or equal to speed of light. What I have written? Hence, when the events are related as cause and effect, the time order cannot be reversed unless the speed greater than c is allowed. Instead of this particular train, you could have taken any other example. And still, you could have seen that the causality will not be violated. And in the case, there are events which are related as cause and effect, then they will necessarily turn out to be time like events. And in that case, time order cannot be reversed. Let us take one more example. This is still a simpler example. And example like this we have been discussing earlier. There is an old example of a ball being thrown in a train compartment of proper length L prime. It means there is a particular compartment which is moving. It is a relative to earth, assuming earth to be a inertial frame of reference. And the length as measured by an observer in the compartment is L prime because that is the proper length because the compartment is at rest in its own frame. One is, ball is thrown towards the front wall and another is thrown towards the back wall. This is what I have shown in the picture, similar picture we have done earlier that this is a person which throws a ball to the front towards the front of the wall. The front being described like this. If this is the direction of the speed, then this wall is the front wall. This wall is the back wall. So, a person throws simultaneously one ball on this side, another ball on this particular side. And my event number one and even number two are corresponding to the events of the ball being received at the two walls. This is what I have described here. Event one, ball reaching the back wall of the train. Event number two, ball reaching the front wall of the train. Now, if we remember this type of problem we have done with ball, then we have done with the speed of light, light being flashed in the two ends we have done many times. Because the ball is being thrown from the center of the compartment, therefore each ball has to travel a distance, a horizontal distance along the x direction as L prime by 2. And both of them travel with the same speed as seen in the S prime observer, as seen in the S prime frame of reference. Therefore, both the walls would reach the wall together. It means delta t between these two events would be equal to 0. Now, as far as the first event is concerned, I assume that the center of the compartment is at the origin. Then the first event occurs at a distance of minus L by 2. The second event occurs at a distance of plus L by 2. So, let me draw my compartment here. This is the center point. A ball is thrown to the right and a ball is thrown to the left. This distance is L prime by 2. This distance is L prime by 2. Therefore, if I calculate this speed I have said U prime. This speed I have said as U prime. If I have to calculate the time that it takes, then all I have to do is to divide this L prime by 2 by the speed. And because this distance is same to this distance, this speed is same as this speed. So, the time when this ball hits here and the time when this ball hits here, these two times must be same. So, if this I call as t1, if this even I call this time as I call as t2, t2 minus t1 must be equal to 0. But this event as far as the first event is concerned, it occurs before the origin at a distance of minus L prime by 2 and at a distance of L prime by 2, but it is in the negative sign. These events occur in the plus x value at a distance of L prime by 2. So, if I take this as the difference, the difference will be turning out to be L prime. So, therefore, delta x prime will be equal to L prime. So, this is what I have written here, that delta x prime is equal to L prime and delta t prime is equal to 0, because delta t prime is obviously equal to 0. Therefore, delta x prime is greater than C delta t prime because L prime is positive and nonzero. Therefore, because delta x prime is greater than C delta t prime, these events are space like separated events. And therefore, it is possible that the time order gets reversed as far as these events are concerned, but it is not possible that the position order gets reversed. Let us assume S frame, which is the ground frame. An observer is observing while sitting in the ground, this particular train. I apply the Lorentz transformation. Of course, I have to apply inverse transformation here because my S frame of reference is moving with speed v. So, if I say delta x is equal to gamma delta x plus v delta t plus sign because this is the inverse transformation. Delta x was L prime, delta t prime was equal to 0. Therefore, delta x will turn out to be equal to gamma L prime. I would not discuss this particular issue because we have discussed this particular issue earlier. And as we have said, that is not possible to find that this particular x delta x is 0, but in this case, I have taken a specific example. Now, if I take delta t, delta t is equal to gamma delta t prime plus v delta x prime divided by c square. Delta x prime is L prime. This is 0. So, delta t turns out to be equal to gamma v L prime by c square. So, we can see that the time difference S in S frame of reference will turn out to be v gamma v L prime divided by c square. Now, of course, depending upon v, if v happens to be positive, this delta t will turn out to be positive. If v happens to be negative because I can always assume that this particular train was moving the other way, that this delta t will turn out to be negative. Therefore, depending upon the sign of v, delta t can be positive or negative as we have seen in this particular example. Now, all I would like to emphasize that these two events, though they appear to be thrown by the same person from the center of the box at the same time, these are not cause and effect related events because the motion of the first wall is independent of the motion of the second wall. It is not because that the wall, one was thrown that wall 2 has to be thrown. It is just a matter that we decided that person sitting there decided to throw the two walls at the same time. Otherwise, he could have thrown one wall earlier than the second wall. In no way, the motion or throwing of the first wall is related to the throwing of the second wall. Therefore, as far as these two events are concerned, they are not cause and effect related. Though it appeared as if the same man is throwing the two walls. So, maybe one is an outcome of the cause of the second event. This is not correct because these two walls being thrown are sort of independent of each other. Hence, it is possible to revert the time order. But I could have gone into a slightly different situation when I could have made these two events later. Let us just see. Let me first read whatever I have said. Note these events are not cause and effect type. If there was a single ball thrown and the events would have been related to the motion of that single ball, then they would have been cause and effect related. And then only they will be time like separated. Just take an example here. Suppose this person would have been sitting at the end of the compartment and was throwing a ball from this particular end towards the front. And event number one would have been throwing the ball. And event number two would have been received, ball being received at the front end. Then these two events are cause and effect related. Because unless this ball would have been thrown, the ball would not have reached here. And in this case, we will find that the events are time like. That is what I wanted to emphasize. So, let us look at the situation and assume that event number one is ball thrown from the back wall. And event number two is ball reaching the front wall of the train. I can find out exactly the same way. What are delta x and delta t corresponding to these two events? Of course, delta x is l because the first event occurred at the back of the compartment. The ball was being thrown here. So, first event occurred here. Let us assume that my origin is now here. So, the first event occurred at this particular point. Second event occurred at this particular point. These two are separated by distance of l. So, therefore delta x as seen in the observer s in the compartment frame is obviously l. So, delta x has not changed, which is l. But now the ball takes a finite amount of time to go from this particular end to this particular end. So, now delta t is not equal to 0. Delta t is equal to the time taken by the ball to travel a distance from here to here in the x direction. And that distance will be given by l prime divided by the speed. So, this is what I have written here in this particular transparency. The delta x prime though is equal to l prime, but delta t prime is now l prime divided by u prime. Because this is the time that this ball will take to go from the back end to the front end as seen in observer s prime. And if I calculate delta x prime minus c delta prime, I have to multiply this by c. And if I multiply by the c and take l prime common. So, I can write this, I take this l prime out. So, in the bracket, I will be left with this l prime has been taken out. So, this will be 1 here, l prime has been taken out. So, it will be 1 divided by u prime. I have multiplied by c. So, it will be c divided by u prime. And because c is expected to be larger than u prime, therefore, this quantity is negative. Therefore, c delta t prime is larger than delta x prime. And these two events become time like separated events. Then it is not possible to find a frame of reference in which the time order of these two events will get reversed. Now, let us try to calculate delta t just to see that this cannot become negative. So, let us let me come back to the frame s. Apply inverse Lorentz transformation. I write delta t is equal to gamma delta t prime, which was l prime divided by u prime as we just now discussed. Plus, because this is an inverse transformation, then v delta x, v is anyway here, delta x was l prime divided by c square. So, this is what will be the value of delta t as seen in s frame of reference. Now, what I am doing? Gamma is anyway out of this particular bracket. I take l prime also out of this particular bracket. And I also force u prime outside this particular bracket. So, I take l prime divided by u prime common from this particular bracket. If I take here, then this factor will become 1. If I am taking l prime, anyway l prime is here, but there is no u prime here. So, I have to multiply it by u prime. So, this factor becomes equal to gamma l prime divided by u prime multiplied by 1 plus v u prime by c square. As you can see that this delta t cannot be negative. Of course, it cannot be negative if v is positive. But even if v is negative, this particular factor because v and u prime both are going to be smaller than c individually. Therefore, this factor cannot exceed c square. Therefore, this factor has to be always less than 1. And even if it is of negative sign, the whole quantity delta t will always lead me a positive value. Hence, what I have written, delta t cannot be negative irrespective of the sign of v. So, long the speeds are lower than c. And therefore, it is not possible to find a frame in which the time order gets reversed. This is what was sort of discussion or giving some examples about time like and space like events. Now, let us go little ahead. Is Lorentz transformation enough? Has he given me all the information about relativity or do I require something more? What I would like to mention here is that this was only the first step. Just doing Lorentz transformation is not enough in order that the postulate of special theory of relativity that all the laws of physics remain same in all the frames. It is necessary to do something more. And what I would like to say is that at least I would have to redefine my momentum. So, now I will give you a very simple example. It is a classical mechanics. This particular example is solved in a very, very simple fashion which we call as completely inelastic collision. In a completely inelastic collision in classical mechanics, we always say that the momentum has to be conserved, but the mechanical energy is not conserved. Actually, if we take the total energy that should be conserved there also, but it is converted into some sort of energy which is not really mechanical. And traditionally in classical mechanics, whenever we are solving collision problems, we apply conservation of momentum of course and conservation of energy which is happens to be conservation of mechanical energy only in those cases where the collisions are elastic. It is possible to have nonelastic collisions or the example which I gave you as completely inelastic collision in which the two bodies come together and get stuck to each other in which we do not conserve the mechanical energy. We conserve only the momentum. Now, we take this particular example of completely inelastic collision as given in a particular frame. And of course in this particular frame, we will say that momentum is conserved. Now, what I would like to show that if I change my frame of reference to a different frame of reference, then in that particular frame of reference momentum need not be conserved if we just take only the Lorentz transformation into consideration. Therefore, something else is need to be done to make conservation of momentum which we believe to be one of the fundamental principles of physics to be valid in all inertial frames. So, I take one specific example. The example is essentially very simple. But even in this particular example, we will show that this violates the universality of conservation of momentum. So, this is what example I have, title I have given need to redefine momentum. Consider a completely inelastic collision in S frame. So, let us assume that there is a frame S in which one particular particle of mass m is found to travel to the right with a speed of 0.6 C. In the same frame, another mass travels to the left with same speed which is 0.6 C. Now, the observer S finds that these two masses collide and then get stuck to each other. This is what we call completely inelastic collision. So, this was a situation before collision where one mass was travelled to the right with a speed of 0.6 C. Another mass exactly identical was travelled to the left with speed 0.6 C. They have a head on collision and then the two masses get stuck together. So, what is remaining is just a single mass of total mass 2 m, the two masses stuck together. And this particular mass comes to rest. And as we can very easily see that this will conserve momentum because if I take the momentum of the first particle that will be given by m into 0.6 C. If I take the momentum of the second particle, the magnitude of that will also be m into 0.6 C. But because the speeds are in different direction, opposite direction, therefore the momentum direction, momentum being vector will be opposite. So, this will be the net momentum because this magnitude is same as this magnitude. So, this will be equal to 0. So, this will be what I call as the initial momentum. It means according to conservation of momentum, the final momentum must also be equal to 0 because mass is not 0 obviously. Therefore, it means the mass must come to rest because then only it is possible for momentum to be conserved or to become 0. So, the speed has to become 0. And that is what I am seeing that after the collision the two masses have come together and got stuck have come to rest. Therefore, the final momentum is also 0. So, we have a situation where initial momentum is also equal to 0, final momentum is also equal to 0. Therefore, momentum is conserved. So, as we have seen just now that the initial momentum is 0, the final momentum also is 0. Therefore, in S frame the momentum is conserved. Now, let us try to look this particular collision from one particular frame, one expressive frame S which happens to be moving with the same speed as the initial speed of the first particle which is 0.60. And we show that in this particular frame this conservation is violated. Even if we can show violation in one particular frame of reference the law gets violated. So, just one example contrary is enough to say that this particular law is violated. So, let us go to S frame of reference in which the relative speed v is 0.60. As we have seen the standard formula for velocity transformation is u x minus v divided by 1 minus u x v by c square. In this particular case the relative speed between the frames has been chosen as 0.60. If we apply this particular thing to the first particle that first particle has a u x which is equal to same as 0.60. I substitute these two quantities in this particular formula to find out the speed of this particular particle in S frame of reference, which is what I have done here. So, u x was equal to 0.60, v was equal to 0.60 divided by 1 minus u x into v divided by c square, c square gets cancelled this becomes 1.36. But as you can see numerator we have 0.60 minus 0.60 this gives me 0. So, irrespective of the denominator of course denominator is not equal to 0. I will get for the first particle the x component of the speed to be equal to 0, which is expected. Because if a frame of reference is moving along with the particle with moving with the same speed as the particle that particular in that particular frame of reference the particle would appear to be at rest would not be appear to be moving. Therefore, I am getting the speed to be 0, which is obvious. Now, let us calculate the velocity of the second particle before the collision. This particular particle was having the same speed, but in a different direction. So, my u x was equal to minus 0.60. So, therefore, I write here as minus 0.60, v was anyway 0.60 and because this is negative. So, this particular sign becomes plus. So, I get 1 plus 0.36, which when I calculate is turning out to be minus 1.2 divided by 1.36 c. So, observer s prime would notice that the speed of the second particle of course is in minus x direction and is equal to 1.2 divided by 1.36 c. Now, let us try to calculate the speed of the two particles, which are of course now clubbed into a single particle after the collision. This particular particle was at rest in s frame of reference. So, let us try to calculate this speed here. The speed in s frame was 0. Here v is 0.60. Therefore, which I am saying is the final speed u f x prime is equal to 0 minus 0.60, which is v divided by 1 minus u which is equal to 0, v which is 0.60 divided by c square. This quantity becomes 0 because of 0 here. Denominator is 1 and you get minus 0.60. Therefore, the speed of the two particles to combine particles after the collision as seen in s frame of reference will be turning out to be equal to minus 0.60. Now, let us try to calculate the moment of the particle, the initial and the final momentum. The first particle in s frame of reference was having a 0 velocity. So, obviously it has a 0 momentum. The second particle was having a speed of minus 1.2 divided by 1.36 c. This multiplied by the mass would be the momentum of the second particle, which happens to be the total initial momentum as observed in s frame of reference. This is the initial momentum as observed in s frame of reference, which will be equal to minus 1.2 divided by 1.36 mc. Now, final momentum, there is only one particular particle which has a mass 2 m and that particular particle as we seen in the previous transparency was moving at the speed of minus 0.60 in s frame of reference. So, this multiplied by the mass of the particle which is now 2 m, the final momentum will turn out to be equal to minus 1.2 mc. As you can clearly see that this is not same as this. So, according to s prime observer, the momentum of the particle has not conserved. Though according to an observer in s frame of reference, this collision really led to a conservation of momentum. But if everything whatever I am saying is true, then according to an observer in s frame of reference, the momentum was not conserved in this particular process. Hence, either we say that conservation momentum is not a fundamental principle of physics, which I am not sure anyone of us will agree. It means that we must change, we must do something else so that the momentum is also conserved in s frame of reference. Therefore, we need to redefine my momentum. So, in the event, we will summarize what we have discussed today. We discussed two examples related to space and time like separated events. And specifically, we saw the cause and outcome related events. Then finally, we discussed one simple example of completely inelastic collision and we said that we must redefine our momentum. Thank you.