 In our last lecture, we had worked out a long example. The idea of working out that example was just to give you some feeling of how multiple events problems have to be solved. And we had discussed specifically the importance of identifying the events and filling the event table. So, in most of the relativity problems involving kinematics, it is important to realize what are the events and then write the coordinates of these events including time. Time of course, we will be treating as coordinate as we have been telling and then fill the event table. Of course, event table can be filled with whatever information is been given in a given frame. If the information is not available in that frame, but in a different frame, that information has to be translated back to a given frame by using an appropriate transformation. So, that is what is the basic trick of working out the problems of relativity. Look at the events, fill in the event tables, find out what are the information available in your frame, whatever are not available in your frame, what are available in a different frame, bring them back to your frame by using an appropriate transformation. That is the way to eventually work out these problems. Now, in today's lecture, we will start with giving one more example involving the speed of light before we go a little bit ahead in the course. So, example is comparatively simple. We have been working out these type of examples when there is a train in which a particular observer is sitting and that particular observer emits light. But in all earlier problems, the direction of the speed of the light or direction of the velocity of the light was always along the motion of the train or opposite to the motion of the train which we have called as x axis. So, C was always pointed out, was always pointing out either in plus x direction or in minus x direction. Now, we will work out a different examples in which the speed of light, in which the light will be pointing out in the y direction. That is the direction which is perpendicular to the x direction and towards the top. So, this is what is the example which I am just describing. An observer in a train which we are calling as a frame S, flashes a light in y prime direction from floor which gets reflected back from the roof at height of H. So, there is a roof at the top which is at the height of H. I will show you just the figure just now. And this received back, the light is reflected back from the top and received back at the floor of the compartment of the train. Now, what we have to do is to describe the motion in S frame which I am calling as a ground frame or the frame in which this particular train is moving. And we assume of course, this the train is moving with a constant velocity because we are always in the realm of special theory of relativity. So, in our case, the velocities are always assumed to be constant. So, this is a problem which is not really essentially a problem, but a way of explaining various concepts once more with a slightly different type of example. So, this is the figure which I have drawn here. Of course, there is one particular person with a light source in the hand. And let us assume that this light source is right here at the floor of the compartment. This is the train compartment which is moving to the right with a velocity V S seen by an observer S. And as we know that this direction we always call as X direction. Let us assume the Y direction is along the top. So, this particular light is being flashed from the floor. The light goes up S seen by S observer vertically up and then comes back. This is a situation which is as observed by an observer S in this particular train. Now, question is that there is an observer sitting on the ground. In this ground frame which I am calling as S frame, this train is moving with a velocity V. How this particular person would perceive this particular light's motion? That is what is the, in fact, we have worked out many type of similar type of problems in classical mechanics earlier. The persons who are familiar with classical mechanics must be very familiar that a particular person sitting in the train throws a ball vertically upwards and the ball comes back and deflects it back. And same of the events or same experiment is being seen, visualized by an observer sitting on the ground. How their perception will become different? Now, everything is essentially similar except that instead of ball, we have now a light and because we are talking of light, so we have to talk with special theory of relativity and use the relativistic transformations. Otherwise, the problem is somewhat similar. So, let us define our events. We have emphasized quite a bit on the importance of identifying events. So, we have event 1. Event 1 is light emitted from the floor of the compartment or the train towards the roof. So, light is thrown upwards. So, this is event number 1. The light being emitted from the roof, from the floor of the compartment and train to go towards the roof. Then, the light goes up, comes back and received back at the ground. So, when it is received back, we call that as event 2. So, my event 2 which I have written here, E 2 is light reaching the floor after reflection from the roof. So, light being thrown, even number 1, light goes up, comes back, received at the ground, that is event number 2. Now, let us first look how an observer in S prime would observe because the experiment has been described with respect to an observer S prime which is sitting in the train. Now, observer in S prime will always feel that light is always going vertically upwards. So, initially when it is going upwards, then the direction of the velocity is along the plus y direction. Once it goes up and gets reflected back, the light starts moving downwards and the velocity now is downwards in minus y direction. That is what will be the perception of the observer sitting in the train. So, this is what I have shown in this particular figure. This is with reference to the observer in the train which I am calling as the frame S prime. It has been given in the example that this particular height is H prime. This arrow shows the direction of the velocity initially, velocity of light initially which goes upwards. This arrow shows when the light is reflected back and tries to come back towards the floor. So, initially the light moves upwards which is the plus y direction, plus y prime direction to be more precise and then it comes back to the floor which we will call in minus y prime direction. So, now let us try to write the events in this particular for these two events in S prime frame of reference. So, let us assume that light is emitted from the origin. So, at the point when the light was emitted, that point was the origin of S prime. In this particular case, z axis is not involved because we are looking only at the x direction and the y direction. So, I am not writing the z coordinate. But because this is the origin, so I can write that x prime is equal to 0 and y prime equal to 0. Let us also assume that time according to S prime observer was also 0 when light was emitted. So, it is easy to write the coordinates of event 1 which is x prime is equal to 0, y prime is equal to 0 and t prime is equal to 0. So, the light was emitted from the origin of S prime and when light was emitted at that time the watch of the observer S prime was showing a time equal to 0. Now, even 2 as we have discussed is that light goes upwards and comes back. So, light goes upwards. This distance is H prime. Then light comes backwards. Again, it travels a distance of H prime. Remember, every information has been given in S prime frame of reference. So, the total distance that light travel will be H prime while going up and H prime while coming down. So, total distance is 2 H prime and we know that light travels with the speed of light. So, therefore, if I have to find out the time taken by the light to go up and come back will be the total distance which is 2 H prime that the light has travelled divided by the C, the speed of light. So, it means the second event must have occurred at a time which is 2 H prime divided by C. When the light is received back at this particular point, remember that point is again at the origin because if light goes vertically upwards and comes down, it is received back exactly at the same point. Therefore, X coordinate and Y coordinate of the second event has not changed because both these events occur exactly at the same point. The light is emitted from this point, goes up, goes up and comes back at this particular point. Both these events light being emitted and light being received occur at the same point which is the origin of S prime frame of reference. But the second event occurred later than the first event because light took a finite amount of time to go up and come down. So, the second event must have occurred at a time 2 H prime divided by C. So, this is what I have written here in this particular transparency that for the event number 2, X prime is equal to 0, Y prime is equal to 0 because the event does occur at the origin and it occurs at time t prime is equal to 2 H prime divided by C. So, these are the coordinates of the event as seen in S prime frame of reference. Now, let us look at the velocity components. Again, with reference to an observer in S prime frame of reference, when the light is moving upwards, we have all that direction as a plus y direction. So, the light had only a y component upwards. When the light was travelling down, again it had only a y component because light as seen in S prime frame of reference is only moving either in plus y direction or in minus y direction. It has no component of velocity in the x direction. That is what an observer sitting on the train sees that light goes vertically upwards, vertically downwards. So, light never moves in a horizontal direction or in any other direction. So, obviously it means that u x prime must be equal to 0 because it has no component, no x component of the velocity and it has no z component of the velocity. So, u z prime must also be equal to 0. It is only u y prime which is not equal to 0 and this is plus c while going upwards and it is minus c while going downwards, while coming downwards. So, I have written u y prime is equal to plus minus c where the signs have to be interpreted appropriately by choosing plus sign whenever I am referring to the motion of light going upwards and I had to choose a minus sign when the light is coming downwards. So, these are the velocity components of the light as seen in s prime frame of reference. Now, my next question is that what will be the velocity component as seen by an observer of the ground which I am calling as s frame. Remember, it is in this ground frame that this train is moving with a velocity v. Now, remember all the information has been given in s prime frame of reference and if I have to find out the information s frame, we have to use a inverse transformation. So, what I would do? I would use an inverse velocity transformation to find out what will be the velocity components of the light as seen in s frame and sorry as seen in s frame of reference which is the ground frame of reference. So, let us write the velocity transformation equations. Velocity transformation equations were u x prime. In fact, I will write u x is equal to u x prime plus v 1 plus u x prime v by c square. This is as far as the transformation of x component is concerned. Remember, we have used a plus sign here and I have used a plus sign here because this is an inverse transformation. Similarly, I will have u y is equal to u y prime divided by gamma 1 plus u x prime v divided by c square. I am not writing for the z component because as far as z component is concerned, this is sort of irrelevant as far as this particular example is concerned. So, let us use this particular example, this particular equations, transformation equations and try to find out u x and u y and to see what are those values as it will be seen in s frame of reference. So, let us look at the equation. We have u x. We had u x prime which was equal to 0. Remember, in s frame of reference there was no x component of the velocity. So, this is 0. v is the relative velocity between the frame. So, it is anyway present here divided by 1 plus u x which is 0 multiplied by v divided by c square. So, this particular thing gives you 0. So, in denominator you are left only with 1 and this becomes 0 plus v divided by 1 which is just equal to v. So, u x turns out to be equal to v which is the relative velocity between the frame. Let us look at u y. This is plus minus c. Again, we have said that the signs have to be appropriately interpreted plus will be when it is going upward minus when it is going downwards. So, I have just put in one equation plus minus c divided by gamma multiplied by the same factor here which we have anyway discussed is 1. So, what you get is plus minus c divided by gamma. So, the y component of the velocity of the light as seen in s frame of reference is plus minus c by gamma. Of course, u z is anyway 0 because u z prime becomes equal to 0. Remember, the transformation equations of y and z component look appear very similar. So, what we have done is now found out the velocity components as seen by an observer in s frame. We just note that according to an observer in s frame, the light is also travelling in x direction. It has also a component in x direction. So, again similar to what we have observed classically that if a ball is thrown vertically upwards as seen by s prime, this ball does not really move vertically upwards as seen by the observer here, but it also develops x component of the velocity which is same as the velocity of the train. That is what we have seen in classical mechanics also. That is what we are seeing here also that this particular light now has developed x component and this x component is same as the velocity of s prime as seen in s which is the same as the velocity of the train. But now, in contradiction to the classical mechanics, we have found even the y component has changed and the y component is now plus minus c by gamma. Remember, this has to be done because as seen in s frame also the speed of light has to be same which is c. So, if it has picked up a velocity component in x direction in the y component, the velocity component must come down in order to maintain the speed of light to be c also in s frame of reference. That is what these transformations have to do because that is the first postulate of that is the postulate of special theory of plating. Now, let me just first check whether I am really getting the velocity of light to be c as seen in s frame of reference. Let us first do a quick exercise because that is what I expect. y component has come down, s component has been picked up, it is from 0 it has gone to v, y component has come down. But I expect that the speed still the speed of light must remain c in s frame of reference. So, let us verify it. So, that is the half set. First check if the speed of light is still c as seen in s as expected. Now, this is what we expect as per the postulate of special theory of relativity. So, when I calculate the magnitude of the speed of the light, I will use this particular standard formula u is equal to under root of u x square plus u i square plus u z square. u z is anyway 0. So, let us ignore this particular thing. We have seen that u x is just equal to v, the relative velocity within the frame. So, for u x square, I have put v square. For u y square, we have just now seen that u i was plus minus c by gamma, where the signs have to be appropriately interpreted. But once I square it, whether it is plus sign or a minus sign, both will make it plus. So, this particular equation would just reduce to v square plus c square by gamma square. Now, what I do is just substitute the value of gamma square. As you remember, gamma square will be equal to 1 upon 1 minus v square by c square, because this gamma square is in denominator. So, this 1 minus v square upon c square will come into the numerator. That is what I have written in this particular transparency. This is v square plus c square by gamma square. And this particular gamma square has been written as, this c square is still here. And this gamma square has written, been written as 1 minus v square by c square. If you expand here, you will get v square plus c square. And when c square gets multiplied to this particular factor, this becomes minus v square. So, you will have v square minus v square that will cancel giving you just plus c square and u becomes equal to c. In fact, it becomes plus or minus c and depending upon the way you want to look at it, but I am looking only at the magnitude. Therefore, I am putting only the plus sign. So, what I have shown is that the speed of light indeed turns out to be equal to c as seen in observer s as expected from the second postulate, from the postulate of special theory of relativity. So, this is what I have written. Now, let us find, now let us find the way the observer in s would find these events. Now, let us look and how the observer in s would find these events. Now, according to observer s, the light has picked up a x component of velocity. It means light is not going vertically upwards, but it is going at an angle. And the angle will depend on the value of gamma of course, because ui depends on the value of gamma. So, if the observer in s frame wants to look at the motion of the light, this is what he would probably visualize. He would see that the light has an x component and also has a y component and is not emitted vertically. What I have written? According to s, the light was not emitted vertically, hence would not fall back at the same coordinate where it was emitted. Because the light does not go upward, but goes like this, then when it falls back, it has to fall at a different point in x. So, according to him, the x coordinate must have changed for this, for the second event. So, the interpretation of this particular observer in s would be that the second event has occurred not at the same value of x, but has shifted. And shifted by how much? It depends on how much time the light took to go up and come back as seen by observer in s frame. Then whatever was the x component of the velocity in that frame multiplied by that particular time. Let me just explain this particular point little more. So, according to s observer, the light is moving at an angle like this, it comes back and does not hit at the same point, same coordinate as it has started. So, light is going to go like this and come back like here. If I had to find out what is the displacement, this displacement would depend on the x component of the velocity of light, which is v and the time taken by the light to go up here and come back here. As seen by s observer, if this time interval is delta t, then this distance must be equal to v delta t. Because v is the x component of the velocity and delta t is the actual time taken for the light to go up and come down. So, this is what I have written. The final displacement of the light in x direction would be given by v delta t, where delta t is the time taken by the light to go up and come back in s frame. And this displacement happens to be same as the displacement of that particular point of the train. Because now if the observer s sees this particular point, this particular point also is moving with the same speed, which is relative to the speed of the train. So, this point also moves exactly by the same distance v delta t and the second event occurs also at the same, shifted by exactly the same coordinate. So, according to the observer in s frame, this point also moves by distance of v delta t and the shift in the x coordinate of the two event is also equal to v delta t. That is what the observer in s frame would see. So, I have put this particular thing in this picture. Remember this height h is going to be same in the two frames s and s, because as we have discussed in one of the earlier example, there is no change in the y component. The length contraction occurs only in the x component. So, as far as the y coordinate is concerned, because y prime is always equal to y in Lorentz transformation. So, as far as going up and going down is concerned, the heights are seen to be same. So, though I am writing here h to make it precise that this is being seen by an observer in s frame, but we know that this h has to have the same value as h prime, because this distance is along the y direction. So, an observer in s frame sees that light goes this way, comes back like this and the x difference in the x coordinate of the two events will be v delta t. It does not occur at the same point or same coordinate from which it was thrown. The second event occurs at a different coordinate. This is what will be the perception of an observer in s frame. Of course, this particular point from which the light was thrown that also moves by the second amount. So, as far as an observer in s is concerned, he will also feel that the same point on train from which the light was emitted is received back also by the same at the same point, but this point as well as the light both have moved along the x direction. So, that is what I have said. Note that h is equal to h prime as there is no contraction in the y direction. Let us calculate delta t to find out how much has been the displacement again. I can use a inverse transformation delta t is equal to gamma delta t prime. This is plus sign because we are talking of inverse transformation v delta x prime divided by c square. This is the inverse velocity transformation delta x prime was 0 because as far as an observer in s is concerned, the coordinate of both the events were same. Therefore, delta x prime was 0. So, this becomes equal to 0 and delta t prime was equal to 2 h prime by c and because h prime is equal to h. So, I have written this as just 2 h by c. So, this becomes gamma 2 h by c. So, according to observer in s frame, the time interval between these two events is not 2 h by c, but gamma times 2 h by c. You will realize that this is the same as time dilation formula which has to happen because in s frame of reference, both the events occur at the same point. So, the time interval in s frame of reference was a proper time interval between these two events. Therefore, in s frame, this time interval must turn out to be dilated and that is what we are seeing that this time interval turns out to be equal to gamma multiplied by 2 h by c. So, this is what I have written. Note that delta t prime was proper time interval. Hence, one could have used the time dilation formula directly for obtaining this result, but need not have really written the Lorentz transformation inverse transformation. If we are fast, we can immediately realize that this is a proper time interval in s frame of reference. So, in s frame, if we just multiply by gamma, I will get back the time interval at delta t. Of course, it can also be observed in a different fashion. We can also find out delta t prime in a slightly different fashion. If you look from the perception of the observer in s frame of reference, this particular light travelled a vertical distance of h. So, vertical component of the velocity, vertical distance travelled by the light was h and the vertical component of the velocity was plus minus c by gamma. So, if I know the distance, if I know the speed, the vertical component of the speed, I can find out how much will be the time taken. So, the total distance travelled by the light, total vertical distance travelled by the light as seen in s frame of reference was h going up, h coming down. And this is achieved by a velocity, y component of the velocity, which is plus minus c by gamma. Therefore, I can always find out that total time will be 2 h divided by the y component of the velocity which is plus minus c by gamma. So, this gamma goes upwards and this becomes 2 h by c to gamma. Because I am looking only at the total time interval, so this plus minus sign does not make a difference. So, I can also find out this particular time interval delta t by using this particular way of doing it. I could have also done a direct inverse transformation on delta x and obtained exactly the same result which I have shown in this particular transparency. Delta x can also be found directly by using inverse Lorentz transformation. Delta x will turn out to be equal to gamma, delta x prime plus v delta t prime, delta x prime equal to 0, delta t prime be equal to 2 h by c. Again, I remind that h prime is equal to h. Therefore, this will be equal to gamma times v multiplied by 2 h divided by c. We get the same result. So, if we know the delta x value, we can always find out that v delta t is the same result. So, we just see that in s, light travels with the same vertical distance as seen in s prime, but with a reduced component of velocity. In addition, the light has a horizontal component of velocity same as that of frame. This causes the x coordinate of the event 2 to be displaced from that event 1, from that of event 1 as seen in s frame. This is similar to classical mechanics because in classical mechanics also this event will appear to be a displaced value of x. The difference here of course is that in the relativity, the y component of the velocity has changed as seen in s frame of reference to make the speed of light same. In all, the speed of light remains c, though its direction of propagation was different in s. Now, let us look into a slightly different issue. Generally, it is a folklore, almost everyone sort of knows that one of the outcome of special theory of relativity was that speeds greater than the speed of light were prohibited. We always said that information does not or should not travel or could not travel with the speed greater than the speed of light. This was considered as one of the outcome of special theory of relativity. We have not discussed this particular point so far. Of course, there is one particular point which is mathematical, which as you can see very easily that almost in every expression that we are using transformation, the gamma appears which is equal to 1 upon under root 1 minus v square by c square. So, if v really happens to be larger than c, then this particular factor would turn out to be larger than 1. Therefore, this when I take under root, the gamma will turn out to be imaginary. When gamma turns out to be imaginary, almost all the transformation that I am talking the x transformation, t transformation, everything there may be an imaginary number and I do not really know how to handle all those imaginary numbers or how to physically interpret. So, there is a mathematical reason that if we had really v less than c, probably we will be comfortable because we do not have to be any bother about it. This is one of the simplistic way of looking into this particular thing. But there is another important physical reason of why we think that speeds greater than speed of light should not be possible. And let me spend some time now to explain this particular aspect and this I will explain by using what I call as the order of events. So, let me first explain my basic things before I come to this particular concept of speed greater than or speed being greater than or being smaller than speed of light. So, let us consider two events, any two events and these two events, let us suppose now I am writing the full expression. Let us call event number 1, e 1 which appears at a coordinate x given by equal to x 1, y coordinate is 0, z coordinate is 0 and it occurs at a time t 1 any arbitrary event. Let us look at the second event. Second event occurs at a coordinate x 2 y and z I am taking the 0 to make things simple and occurs at a time t 2. Now, for simplicity let us assume that x 2 minus x 1 is positive which I am calling as delta x. So, delta x I am defining x 2 minus x 1. Similarly, let me assume that t 2 minus t 1 is positive, t 2 minus t 1 I am defining as delta t. So, I am assuming that delta t and delta x both are positive. Remember, both these events are being seen by a particular observer in a frame s. And two events have occurred in such a fashion that even delta x is positive, delta t is positive, delta y delta z is equal to 0, I do not bother about them because y coordinate and z coordinates are assumed to be 0 or to be same or to be precise and I have to take 0. So, let us take 0. Now, my question is that if delta x is positive and delta t is positive, is there a frame in which delta x can become negative? If of course, if you go to different frame I should call it delta x prime, delta x prime has become negative and delta t prime has become negative, one or both of them have become negative, is it possible? So, let us discuss this particular aspect little bit more in detail that if delta x is positive and delta t is positive, is it possible to have a frame in which delta x can change sign? It means delta x prime becomes negative or delta t can change the sign because delta t prime becomes negative or both of them can become negative, is it possible to have any such frame? So, this is the question that I am trying to understand and trying to answer this particular thing. So, let us look at the Lorentz transformation. The Lorentz transformation I have applied in the differential form, this we have discussed even earlier that rather than writing x, I mean I can have used x 2 prime, x 1 prime that take the difference, but I am just taking writing in the differential form. So, I am writing delta x prime and delta t prime directly I am transforming and this is given by the Lorentz transformation which is delta x prime is equal to gamma delta x minus v delta t. Similarly, delta t prime is equal to gamma times delta t minus v delta x divided by c square, exactly the same as Lorentz transformation. I repeat the question can delta x prime and or delta t prime be negative? Now, let me divide this particular problem into two different cases. One I will call case one, another I will call case two and I will separately look into this case one and case two and trying to answer this question is that if delta x is positive, delta t is positive, can delta x prime, delta t prime be negative? So, let us look at the case one which is delta x is less than c delta t. So, we specifically are looking at one particular case where delta x is less than c delta t that is my case one. I have rewritten these equations delta t prime equation I have slightly rewritten I have taken c out of that particular bracket. So, this becomes gamma upon c because I have taken the c out of the bracket. So, there is a c appears here. So, this becomes c delta t remember if I expand then this becomes delta t and there was c square here. So, I have taken one c out here. So, this becomes v delta x divided by c. So, same equation I have written just to make things little simple to interpret I have taken this c out of this particular bracket. So, this is the same equations which I have written in a slightly different fashion. Let us take a specific example. Let us assume that delta x is equal to 0.9 times c delta t. So, delta x is less than c delta t because delta x is only 0.9 times c delta t. Now, let us assume that v is equal to 0.91 c which is slightly larger than this particular factor of 0.9. So, I have chosen one particular value of v which is 0.91 c slightly larger than this factor of 0.9. Let us look and substitute in this particular equation. Let me write this particular equation again delta x was equal to gamma delta x minus v delta t. Now, delta x I have taken as 0.9 c times delta t. We have taken as 0.91 times c this becomes c into delta t. Let me substitute for delta t, gamma divided by c into c delta t minus v delta x by c. I put it here in the transparency gamma by c, c delta t minus v which I have taken as 0.91 c that c cancels it down c and this particular delta x was 0.9 times c delta t. Or you can very easily see that if I take this c delta t out here out of the bracket then this becomes 0.9 minus 0.91. So, delta x has become negative. Look at delta t. If I take c delta t common here this becomes 1 minus 0.91 multiplied by 0.9. When I multiply 0.91 by 0.9 I will get a factor which is less than 1. Therefore, this delta t prime is still not negative is positive because here you have 1 minus something which is less than 1. Now, instead of 0.9 I could have used even a larger value. But still I can always find one particular factor here which is slightly larger than this to make delta x prime negative. But because this factor which I have used here is less than 1, whatever value of v I use here, whatever this particular factor I will use here I will always get this product to be less than 1 until this factor what I am using here the ratio between v and c exceeds 1. So, remember here we have something which is less than 1. In order that if I want to make this more than 1 this factor should definitely be exceeding 1. Otherwise if I am you multiplying this by a factor which is still less than 1 this factor will always be turning out to be less than 1. And remember delta t prime can become negative only if this factor becomes greater than 1. And in order to that this factor becomes greater than this greater than 1 and this factor here is less than 1 this factor here must be greater than 1. So, let us try to consolidate what I have said delta x prime is negative but delta t prime is not. In general for case 1 it is always possible to find a frame in which delta x prime is negative even if it was not 0.9 it was 0.92 then I could have used v is equal to 0.93. But I will never be able to make delta t prime negative unless v becomes larger than c because only in that particular case the ratio between v and c will exceed 1. And then when it is multiplied by a factor which is less than 1 can yield you a value which is greater than 1. Two factors both of which are less than 1 when they are multiplied they will never give you a value greater than 1. Therefore, in this case delta t prime will never be able to be negative irrespective of what we use what v you use unless v is larger than the speed of light. But as far as delta x prime is concerned you can always choose an appropriate value of v which is less than c and can still make delta x prime negative. Such type of events are called time like separated events. We say that these events are separated as time like time is so strong it is a pure time interval that time cannot be reversed. The position can be reversed between the two events I can always find a frame of reference in which the position of the events can be reversed delta x prime can become negative. But I will never be able to find out a frame in the delta t prime changes the sign unless of course the velocity of the frame of reference exceeds speed of light. Now, let us look at the case 2. Case 2 is the case when just suppose delta x is greater than c delta t. I write exactly the same equations delta x prime is equal to gamma times delta x minus v delta t. I write delta t prime is equal to gamma by c c delta t minus v delta x by c exactly the same equations which I have used for case 1. But now because delta x has been assumed to be larger than c delta t. So, let us take one example delta x assume it to be 1.2 times c delta t. So, delta x has become larger than c delta t. We still take v is equal to 0.91 c and let us see what happens. We substitute the value delta x is equal to 1.2 c delta t here and for v I substitute 0.91 c. Similarly, here for delta x I choose 1.2 c delta t and for v I choose 0.91 c. I put it in this equation here if you take c delta t common out of the bracket you will find that here you have 1.2 and here you have 0.91. This of course is positive. If you look at delta t prime you have c delta t you have 0.91 multiplied by 1.2 multiplied by c delta t. If you just multiply 1.2 by 0.91 you will get 1.092. This factor is larger than 1. So, if I take c delta t common outside the bracket you will get 1 minus 1.092 which makes it negative. So, delta t prime will turn out to be negative. So, my conclusion is that in this particular case delta t prime is negative but delta x prime is not. Now, when I say delta s greater than c delta t it could be I could the factor could have been anything. But remember because this factor is larger than 1 unless v itself becomes larger than c this will always remain positive. Therefore, delta x prime will always be positive. Here instead of 1.2 if this particular factor was not 1.2 but was something less I could have still chosen some particular value because this is still greater than 1 I could have still chosen an appropriate value of v to make this particular factor greater than 1. So, even if it is not 1.2 but 1.1 I will make this factor slightly larger. So, I can still find one particular frame some particular frame of reference with a very large value of v in which this delta t prime will turn out to be negative. So, my conclusion is that in such type of cases you will never be able to find a frame of reference in which delta x prime turns out to be negative unless that frame moves with a speed greater than speed of light. But delta t prime you can always find out some frame of reference which still moves with speed less than speed of light but for which delta t prime will turn out to be negative. So, this is what I have written here in general for case 2 it is always possible to find a frame in which delta t prime is negative. However, it is not possible to find 1 when delta x prime is negative unless v is greater than c such type of events are called space like separated events because this interval is pure space. The space cannot be reversed space is strong it cannot be reversed time interval can be reversed their sign can be reversed but the space sign cannot be reversed. So, these are called space like separated events. Now, we have talked about the signs changing this and that you know what does that mean you know what is the physical significance of that. So, let us come back to that particular physical significance of what is meant by delta x prime changing the sign delta t prime changing the sign becoming negative. So, what is the significance of delta x and delta t changing sign when we go to from one frame to another frame. Let us first look at delta x delta x prime being becoming negative in another frame essentially means that the second event occurred at a value of x which is smaller than the first one. So, remember delta x was equal to x 2 minus x 1. So, if x 2 becomes smaller than x 1 then delta x will turn out to be negative. So, all that it means that in another frame of reference x 2 turns out to be smaller than x 1. That is not very surprising let us just look an example which is a very simple classical example let us forget about relativity let us look classically even classically this is possible to see that delta x could change sign. This figure looks somewhat complicated, but let us try to understand it. Let us suppose this is an event which occur here. This is a frame S in which he sees that this event occurs at a distance of x 1 from him or her and let us assume there is a lightning striking here at a given time and it occurs at a distance of x 1 from him. There is a second lightning striking at further away from him at a time later than this. So, x 2 is larger than x 1. So, delta x was positive because this event occurred later than this event. So, t 2 was larger than t 1 therefore delta t is positive. So, as far as s is concerned delta x is also positive delta t is also positive. Remember I am talking only a classical thing there is nothing of relativity here as of now. There is another person train which is moving with respect to s. When this is moving when event number 1 occurred this particular person was close to this observer s. Therefore, he finds that this event occurs at x 1 prime. Of course, if they are in classical mechanics if they are exactly aligned then x 1 should be equal to x 1 prime, but let us forget about that particular issue. We see that this will occur x 1 prime. Now, second event occurs little later during this particular time this compartment was moving towards the right hand side. So, depending upon delta t it is possible that this event let us say occur let us say 10 minutes afterwards and during 10 minutes this particular train has moved from here to here. When it has moved from here to here the second event occurred only at this distance because he has also moved with respect to s. So, according to s prime observer the coordinate of the second event is x 2 prime while the coordinate of the first event was x 1 prime. So, obviously x 2 prime smaller than x 1 prime depends on what is the speed and what is the time difference between these two events. So, it is always possible that he finds that this particular event occurs closer to him than when the first event occurred. So, according to the observer s prime delta x has changed sign. So, even in classical it is possible that if delta x was positive in s frame delta x prime will become negative in s prime frame of reference. So, there is nothing surprising in delta x changing sign as I change my frame of reference, but what does changing sign of time interval means? It means the second event occurred earlier in a different frame of reference than the first event. So, as far as two lightning comes if the second observer observes that the second lightning occurred first and the first lightning occurred second that is what would mean delta t changing sign. Does it make sense? Now, if we are talking of two events which are totally uncorrelated which has no bearing with each other then if it changes I do not bother. For example, I say event number one a particular train departing from New Delhi railway station and event number two a plane flying off from New York airport after some amount of time this event one and event two are totally uncorrelated. That particular plane and this particular train departing has no relation with respect to each other. The departure of that particular plane does not depend on the departure of the train, but let us consider some correlated events in which there is a cause and effect. Event number one train is starting from New Delhi railway station. Event number two train reaching at Mumbai railway station this is event number two. Now, unless the train would have started from New Delhi railway station it would have never reached Mumbai station. So, these are related events occurrence of second event depends on event number one. Take another example I shoot somebody with a gun and that person dies. Now, unless I would have shot that person would have not died. So, his death which I call as event number two and if I call my shooting the gun as event number one event number two has to occur because event number one had occurred. If event number one had not occurred event number two will not occur. Now, if delta t prime changes sign it means these orders can also change sign. Let us just see I am calling these events as caused events. Reversal of time interval would mean that it is possible to find a frame in which causality can be violated. Let me explain this particular point in little bit. If delta t prime changes sign it means it may be possible that in some other frame of reference one could see a train arriving before it starts or in some other frame one could see that a person dies before somebody has shot him that is what changing sign delta t prime would mean. This is what we call as causality because even number two is the effect there is the cause is even number one because of which even number two occurred. Now, normally we never see in our daily life that causality is violated. I should not be able to for my physics physical idea says that should not be possible for me to find out the frame of reference in which these two events which are one which is a cause of another in which the time interval could change because this would mean that in some other frame of reference a person could be found to die before he was shot at which normally we do not explain. Now, if the events the two events that I am talking the related events are space like events only in that particular case it is possible to invert the time delta t to make delta t prime negative. But in order that events are space like delta x must be greater than c delta t. It means the distance between these two events must be larger than c delta t. It means if I am shooting the gun here and the person is dying at a later distance let us say 2 kilometers away then this distance should be larger than c delta t. It means the bullet must have travelled with speed greater than c delta t. Then only it is possible that the events the correlated events become space like and it is only in space like intervals but delta x is larger than c delta t it is possible to revert time interval. And if the events are time like where delta x less than c delta t anyway it is not possible to invert the time interval unless v is larger than c. So, if we say that if v larger than c are not prohibited it will sustain it will make causality intact it will not be possible for us to see that the cause comes later and the effect comes earlier in any other frame of reference. Therefore, if I say that if I restrict my speeds to speeds less than speed of light the most equal to the speed of light I will maintain the causality. That is what led to the conclusion that speeds greater than the speed of light should not be acceptable in spatial theory of relativity. So, this is what I have said as conclusion speeds greater than c are prohibited causality cannot be violated. And then I will give the summary of whatever we have discussed today. We of course, discuss one example in which we had sort of included the speed of light and taken the speed of light and two in two different frame of reference. And then we have said that in order to maintain causality the speeds greater than c should not be allowed.