 In our last lecture, we had applied the postulates of special theory of relativity and eventually arrived at Lorentz transformation. We do not say that we derive Lorentz transformation, we just say that we arrived at it. We presented a series of arguments on how Lorentz transformation could be arrived at. The real test that Lorentz transformation is believed on is on experiment. If we find an experimental support, then only we believe on Lorentz transformation. After that, we took an example, which was an old example of a light being emitted from the origin and we had earlier found out that under Galilean transformation, the speed of the light does not remain constant if we change the frame of reference, which was obvious because the classical mechanics was not designed for that. Then we applied Lorentz transformation on the same example and said that if we apply Lorentz transformation, then the velocity of light turns out to be same in the other frame also, which is what Lorentz transformation is based on or the second postulate of special theory of relativity is based on. So, that is the way we proceeded in our last lecture. Just to recapitulate, let us write or let us discuss the Lorentz transformation again. These are the equations governing Lorentz transformation. In fact, these are the equations. We had introduced two symbols. One is beta, which is v divided by c. Just to remind, v is the relative velocity between the frames and c, of course, is the speed of light. Then we defined another parameter called gamma, which was 1 upon under root 1 minus beta square. If we use these two parameters, the Lorentz transformation equations become like this. For x coordinate, its x is equal to gamma multiplied by x minus vt. y prime remains equal to y, z prime remains equal to z and t prime becomes gamma multiplied by t minus vx divided by c square. We had earlier said that in the non-relativistic limit, meaning when v is much smaller than c, this beta is negligible in comparison to 1 and gamma is essentially close to 1. Then x prime just becomes x minus vt, which is the standard Galilean transformation. These two equations are common even in the Galilean transformation. Here, if v is very small in comparison to c, this particular term is neglected or can be neglected. This gamma will tend to 1, then t prime will be tending equal to t, which is again the standard classical Galilean transformation. So, these equations effectively become important only when v is close to c, where this factor beta square cannot be neglected in comparison to 1. Today, we will try to discuss some of the examples, some other examples of Lorentz transformation. Specifically, two very, very important and popular results arrived out of Lorentz transformation. One of them is called length contraction. The length contraction as the name suggests means that the length of an object may not be constant if an observer in a different frame observes the length of the same object. In fact, as we will be seeing there is a contraction or length becoming smaller is associated with this phenomenon, which is called length contraction. But before we really derive it, let us understand a few terms because it is important to derive or to work with length contraction formula correctly. It is important that we are familiar with these things. The first term is the proper length. Proper length as the name suggests or as we have defined of a particular rod is the length measured in a frame in which it is at rest. So let us imagine that we have to measure the length of this particular rod. This is a rod, the length of which we have to measure. Now this particular rod is stationary in my own frame of reference. So I do not see any motion of this particular rod because as I said this is stationary in my frame of reference. So whatever is the length that I would measure for this particular rod would be called proper length. If this particular rod was moving in my frame of reference, then the length that I will measure of this particular rod would not be called proper length. So proper length has a specific meaning that it must be measured in a frame of reference in which this particular rod is at rest. And that frame of reference in which it is at rest, that particular frame of reference is called proper frame of reference. That is what we have written in this particular transparency. Proper length of a rod is length measured in a frame in which it is at rest and proper frame is a frame in which this rod is at rest. Now let us assume that this particular rod is along the x direction. Remember this particular x direction is the direction of the relative velocity between the frames. So let us assume that this particular rod is inclined or is put along the x direction or it is put exactly on the x axis to be more precise. So this is what I have written. Let us first assume that the rod is placed along the x axis. Then we will show just now that its proper length is the largest and if we go to any other frame in which there is a relative motion along the x direction, then its length will turn out to be shorter. That is why it is called length contraction. Now let us suppose that we have to measure the length of this particular rod. How do I measure? The question is very simple that I measure the coordinates of the two ends of the rod. If I measure the coordinates of the two ends of the rod, actually by subtracting these coordinates I can always find out what is the length of this particular rod. So if this rod's length is measured, I can measure what is the coordinate here. I can measure what is the coordinate here. Then take the difference of these two. This will give me the length of the rod. But the question which I am posing is that is time important. Let me explain what I mean to say that suppose I measure this particular length, this particular end or coordinate of this particular end of the rod, let us say at t is equal to 0. Now this end, this particular coordinate can I measure at any other time? Is time important for its measurement? See the answer to this question is that, no it is not important, so long this particular rod is at rest. So I can measure this coordinate now and go and have a cup of tea, come back and then measure the other coordinate. Still the difference of these two coordinates will give me the length of the rod because this rod is fixed in my frame of reference, it is not moving. So this rod is not going to change its coordinate as a function of time. So it is immaterial when I decide to measure the coordinate of this particular end and when I decide to measure the coordinate of this particular end. But suppose this particular rod was moving in my frame of reference, then the same statement is not correct because if I make measurement of this particular end now and make the measurement of the coordinate of this end little later during this particular time, this rod has been moving. So this rod is moving, I make a measurement now and by the time I go to the other end make the measurement, by that time the rod has been displaced from its position. So its coordinate has changed. Therefore if you still I decide to take the difference of the two coordinates that will not give me the correct length of the rod. So this is what I have said, measure the coordinates of the two ends of the rod, it is time important for the measurement, yes only if the rod is moving, then it is important that the two ends of the rod must be measured at exactly at the same time. The situation I have picturized here in the next transparency, I have depicted a moving rod. This red is the position of the rod at a given time, let us call time t is equal to 0. Now let us suppose at time t is equal to 0, I made the measurement of coordinate of this particular end and find out, find it out that it is equal to x1. Now I make some delay, I do not immediately make the measurement of this particular end but after some time I decide to make the measurement of the other end of the rod. Then by this particular time this rod has moved ahead. So at a later time what I would be measuring is probably this coordinate which is x2. So you can clearly see that x2 minus x1 is not equal to the length of the rod. It is a different question that from these two measurements also I can find out the length of the rod provided I take care of this displacement that the rod has taken or the displacement that rod has undergone during the difference of the time during which I have measured that is a different question. That also I will give an example little later but at this moment what I want to do I want to measure the length of the rod by just measuring the coordinate differences. And for that if the rod is moving in my frame of reference it is important that the difference that the two coordinates must be measured in the same time at the same time only under that condition the difference of the coordinates will give me the correct length of the rod. So this is what I have written if the coordinates are measured at two different time their difference does not give the correct length this is what I have said. There may be other methods of measuring the length of the moving rod some would be discussed later as I have mentioned that if I can calculate how much is the rod how much the rod has moved during the time difference that I have taken or the time I have taken in making a measurement of one end and the other end then of course I can find out the length of the rod that I will do little later. But if the coordinate difference of a moving rod has to give me length then for a moving rod the two ends must be measured at the same time that is what I have written. But the coordinate difference of a moving rod would give correct length only if the two ends are measured at the same time this is an important point to note because this is where I find a very large number of students make a mistake they do not bother about time they just decide and they just say that x2 minus x1 equal to the length which may not be correct because if the object is moving in the frame of reference then x2 minus x1 need not give the correct length until these two measurements were carried out exactly at the same time. Now, let us come back to the problem let us assume that this particular rod is at rest in S frame of reference we have already defined what are our S frame and S frame of reference Lorentz transformation is applicable when we impose certain conditions on S and S frame of references which we have discussed earlier that the axis should be parallel and the relative motion should be along x direction so that is anyway have to be valid. But what I assume in this particular case that it is the S frame of reference in which this rod is stationary it means it is S frame which is the proper frame. So, this rod AB which I am calling here is AB is at rest in S frame of reference obviously it means that in the S frame of reference this particular rod is moving I can always imagine that I have put a rod let us say in a train and the train is moving with respect to an observer on ground. So, a person sitting on the train feels that this rod is not moving, but a person sitting on the ground feels that this particular rod is moving remember this x direction is the direction of the relative motion this is in this direction that S frame is moving relative to S. Now, we have earlier agreed that relativity things are simple if we define events generally it is always a good idea to define events. So, let us suppose we define events relating to the measurement of the length of this particular rod and these events I am defining here calling event number 1 and event number 2. So, this is my event number 1 let me just write show the rod here this was my S frame of reference and this rod was at rest here this was my end A this was my end B and this rod is stationary in S frame of reference being also observed by another observer which is an S frame in this S frame this rod is actually moving to the right as the frame of reference S is also moving to the right with respect to S. So, my event number 1 is that I make the measurement of the coordinate of this particular end of the rod and let us assume that this particular measurement is being done by an observer in S frame of reference. So, he makes a measurement of this particular the coordinate of this particular end of the rod this is my event number 1 then he goes here let me not say goes here but he also makes a measurement of the end B of the rod this I call as event 2. Now, as this particular rod A, B is not stationary in S frame of reference. So, in principle he cannot go from A to B and make the measurement the measurement of the coordinate here and coordinate here have to be done at the same time if the coordinate difference of these two events has to give me a correct length of the rod. So, this is what I have written here I have defined my events my event number 1 is observer in S measures the coordinate of A end of the rod and let us suppose this coordinate B X 1 and T 1 normally we have a tendency of now including time also as a coordinate because as we will be seeing going ahead in the course we realize that for many purposes time X as another coordinate. So, because this event number 1. So, I have put a subscript here 1 to this event number 1 occurs at a coordinate of X 1 at a position coordinate of X 1 and at a time coordinate of T 1. Similarly, the observer in S measures the coordinate of the B end of the rod and let us suppose he does it and finds out the coordinate to be X 2 and the measurement is done at time T 2. So, the coordinate of this particular event is X 2 T 2 while the coordinate of the first event is X 1 T 1 as we have earlier realized and discussed this X 2 minus X 1 will give me correct length of the rod only if T 2 is equal to T 1 because it is in S frame that the rod is actually moving. This is what I have written here X 2 minus X 1 is equal to L in S only if T 2 is equal to T 1 which I have said let us assume that this is equal to T rather than writing same symbol. So, I said let us assume that T 2 1 T 2 is equal to T 1 and is equal to T. Then these two coordinates I use Lorentz transformation to find out the information in S frame of reference. Let me write the Lorentz transformation once more here just to remind you it was X prime is equal to gamma X minus V T Y prime is equal to Y Z prime is equal to Z T prime is equal to gamma T minus V X divided by C square. This is what was my Lorentz transformation. Now, let us apply this Lorentz transformation on to the two events. If I apply let us say first event number 2 it means the value of X was X 2 and time T was T 2 I will get the coordinate of that particular event in S frame of reference. So, I will get X 2 prime the coordinate of the second event in S frame of reference. Similarly, if I substitute here X 2 substitute here T 2 I will get T 2 prime which is the time at which the second event occurs according to an observer in S frame of reference. So, this is precisely what I have written here. If you go back to the transparency it is X 2 prime is equal to gamma X 2 minus V T because T 2 I have taken to be equal to T this is what I have written earlier. So, this I have just put it equal to T then T 2 prime will be given by gamma T which is just T 2 which is equal to T minus V X 2 divided by C square exactly the same thing I put the coordinate position and time here for the first event put position and time for the first event I get these two equations. You can very clearly see that as far as S prime observer is concerned T 2 prime is not equal to T 1 prime. See this T is same as T but this X 2 and X 1 there has to be a difference if the length has to have some I mean if the rod has to have some length then X 2 cannot obviously be equal to X 1 therefore this T 2 prime is not equal to T 1 prime. So, what it means that according to the observer in S frame of reference these two events occurred at different time it should not be surprising to us because remember earlier we have discussed that simultaneity is relative I am I shall be discussing I shall be harping on this issue number of times saying that simultaneity is relative. It is in S frame that the two events occurred at the same time hence they were simultaneous according to S frame of reference these two events did not occur at the same time therefore they are not simultaneous. So, according to him the two events occurred at different time would it mean that S 2 prime will not agree that X 2 prime minus X 1 prime is equal to length no that is not correct because in S frame of reference this rod is stationary. So, it is not important whether the two ends should be measured at the same time what is important is that we just make measurement at any given time. So, even if the measurement of one end of the coordinate in S frame of reference was done in different time from the other end the rod being at rest in that particular frame of reference the time the position difference will still give me the current length. So, what I want to argue out that X 2 prime minus X 1 prime is equal to the correct length of the rod in S frame of reference irrespective the fact that T 2 prime is not same as T 1 prime. While in S frame we could not have done a similar argument I have to force the time to be same in S frame of reference only in that case X 2 minus X 1 would have given me the correct length. Now if I want to find out a relationship between the lengths measured in S and S all I have to do is to take the differences. So, I know that X 2 minus X 1 X 2 minus X 1 was the correct length in S frame of reference because these two times are same X 2 prime minus X 1 prime would anyway be correct length in S frame of reference even though the times were different. So, I will say that X 2 prime minus X 1 prime is L prime the length of the rod in S frame of reference and call X 2 minus X 1 as the length of the rod in S frame of reference. So, if I subtract these two you will just realize that this particular term V T here would cancel time being identical. So, you will just get X 2 prime minus X 1 prime is equal to gamma multiplied by X 2 minus X 1 this is what I have written in the next transparency. We note that T 2 prime is not equal to T 1 prime still X 2 prime minus X 1 prime is equal to L prime this is because the rod is at rest in S frame. Measurement time has to be adjusted to be same in S and not necessarily in S frame. I have taken the differences I get X 2 prime minus X 1 prime is equal to gamma X 2 minus X 1 as we discuss this is the length in S frame of reference this is the length in S frame of reference I call this is L prime I call this is L. So, this L prime is equal to this L is equal to L prime divided by gamma which is what I have written in this particular equation as we have seen earlier that gamma is to be always greater than 1 because you remember gamma was equal to 1 upon under root 1 minus V square by C square. So, you are always subtracting some quantity from 1. So, therefore, this particular factor is always going to be less than 1 and 1 divided by something less than 1 would always make a quantity greater than 1. So, gamma would always be greater than 1 therefore, this L would always be smaller than L prime. So, we say length is contracted it is called length contraction. Let us take an example slightly different example in which I do not make the rod along the X direction, but let it have a slightly different orientation. So, in this particular example, I have assumed that this particular rod is lying actually in X prime Y prime plane. I still assume that it is the S prime frame of reference which is the proper frame of reference for this particular rod. So, this rod is actually stationary in S prime frame of reference, but the only difference I say that this rod is no longer along this particular direction, but making some angle you know somewhere here like this. So, my rod is like that AB and this is S prime frame of reference. So, I assume that the end A of this particular rod is at the origin here. I assume that length of the rod is 1 meter. So, effectively in S prime frame of reference, I know the X coordinate of A which is 0 0, S coordinate of B which I can find out very easily. If I know the length and if I know the angle and what I have to do now is to find out the length of this particular rod in S frame that is what is the question. So, let me read the statement of the question or statement of the example. There is a 1 meter rod AB the one which I have said the length has been given to be 1 meter rod and it is kept stationary in S prime frame of reference which is the proper frame of reference. With end A at origin, the way I just now shown in the picture in X prime, Y prime plane making an angle of 60 degrees. So, I know the theta value which is 60 degrees with X prime plane with X prime axis I am sorry. What would be its length in S? If the relative speed between S and S prime is 0.6 C. So, I have given some numbers. So, we can calculate the answer a numeric answer. Also, find the angle that the rod makes with X axis in S. So, according to S observer at what angle this particular rod is inclined with respect to X axis. So, this is essentially the question. The same picture which I had drawn on this particular piece of paper you have S frame of reference this particular rod is at rest with A being at the origin, this end B being here, this angle being given is 60 degrees and this length is being given to be 1 meter. Now, like before I will define the events and we will define the events to ensure that these two events occur at the same time in S frame of reference because it is an S frame that this particular rod is moving. Also, because here Y coordinate is also involved see in earlier example we did not bother about the Y coordinate because Y was 0 in S frame it had to be 0 in S frame of reference. But here because this particular other end of the rod will have a different value of Y than the first end of the rod B end will have a different Y than the A end. Therefore, it is better to talk about the Y coordinate also. So, we will define our events. So, I said like before we have to define events relating to the measurement of the ends of the rod and these events have to occur at the same time in S. Let the coordinates of these events be again like before even is X 1, Y 1, T. So, this coordinate occurs this particular sorry event occurs at coordinate X 1, Y 1 and T according to an observer in S frame. Similarly, event number 2 I have put subscript 2 happens at X 2, Y 2 and at time T these two times I have put same to ensure that actually the measurement of these coordinates will give me the correct length of this particular rod. So, in principle as we will be discussing if I take the difference of this and take the difference of this and square and add and take the under root I will get the length of the rod in S frame. Again, we have to carry out the Lorentz transformation to find the corresponding coordinates of these two events in S frame of reference. The only difference in this particular case is that I already know the coordinates. So, I can write this equation makes not such a great difference, but just to ensure that we understand the situation well. Now, we applied for the second event the Lorentz transformation which is X 2 prime is equal to gamma X 2 minus V T just like in the earlier example of length contraction. Only thing I say that if this particular length was 1 meter this angle is theta then the X coordinate of this particular rod will be 1 cos theta because 1 is the length of the rod the Y coordinate of this particular rod will be equal to 1 sin theta. So, this is the X coordinate this is the Y coordinate this will be 1 cos theta this will be 1 sin theta because theta has been given to be equal to 60 degrees. So, 1 into cos 60 which is half this will be half meter and 1 sin theta will be equal to 2 3 by 2 meters. So, this particular coordinate the Y coordinate would be equal to 2 3 by 2 meters while this X coordinate will be equal to half meter. So, this is what I have written in this particular transparency that X 2 prime is equal to gamma X 2 minus V T which I know is equal to half meter Y 2 prime is equal to Y 2 which I know is under root 3 by 2 meters. Similarly, I can take the Lorentz transformation of the X coordinate which gives me X 1 prime is equal to gamma X 1 minus V T this coordinate is equal to 0 Y 1 prime is equal to Y 1 is equal to 0 because I know in S frame of reference and A is at origin. So, X 1 prime and X 2 prime I am sorry X 1 prime and Y 1 prime both must be equal to 0. Like before we take these two events to be different in S prime their time will be different in S prime, but I have not specifically written in this particular case the equations involving time difference where they the time transformation because as far as the problem is concerned it is asking only for the length it is not asking for time. So, therefore, I have not written the equation relating to time transformation I have just written the X transformation because for the problem that is not important. Only thing I have to ensure that the time of the two events is same in S that I have ensured by putting the two times to be same that is all. Like before I take the difference of these two just like what we did in the earlier case I will find gamma X 2 minus X 1 to be equal to 0.5 because remember one was 0.5 another was 0. So, the difference is just 0.5 similarly Y 2 minus Y 1 was equal to root 3 by 2 because one was root 3 by 2 another was 0. So, this is this and of course for velocity equal to relative velocity equal to 0.6 C I get gamma is equal to 1 divided by under root 1 minus V square by C square which is 0.6 square this becomes 0.36 1 minus 0.36 under root it becomes 0.8. So, this is equal to 1.25. So, gamma is equal to 1.25 I mean this 0.6 gives you very clean number for gamma. So, in many of the problems to make the mathematics simple making the algebra simple now we choose the gamma value which is something 0.6 0.8 that this gives clean value of gammas. So, I substitute these values of gamma here and find the length as I have said earlier the length is given by X 2 minus X 1 plus Y 2 minus Y 1 square look back at this particular rod if I know the coordinate of this particular rod even if they are not 0. So, what I have to take the difference of the X coordinate and the difference of the Y coordinate this length plus this length this length square plus this length square will give me this length square if I take the under root I will get the length of the rod. So, this is what I have written here the length of the rod in S is thus given by X 2 minus X 1 whole square plus Y 2 minus Y 1 whole square we just now seen from the previous transparency here that X 2 minus X 1 is equal to 0.5 divided by gamma and gamma is equal to 1.25. So, I have written here is 0.5 divided by 1.25 square plus here there was no gamma there is just root 3 by 2 because Y prime is equal to always Y if I calculate this number we get under root 0.91 which is approximately equal to 0.95 meter. Hence the length of the rod in S frame of reference would turn out to be approximately equal to 0.95 meter and not 1 meter. I have to calculate the angle and angle again if I have to calculate this angle I have to calculate this distance this distance divide by the 2 this will give me tan theta. So, this is Y 2 minus Y 1 this X 2 minus X 1 divide the 2 and that will be equal to tan theta. So, this is what I have written here in the transparency that tan theta is equal to Y 2 minus 1 Y 2 minus Y 1 divided by X 2 minus X 1 Y 2 minus Y 1 was root 3 by 2 X 2 minus X 1 was 0.5 divided by 1.25. So, 1.25 comes into the numerator if you calculate this number approximately turns out to be 2.165 which gives theta approximately equal to 65.2 degrees. So, according to an observer in S this rod is inclined not at 60 degrees but 65.2 degrees. I hope you would have appreciated one particular aspect from this particular derivation that the gamma factor arrived only in the X component and not along the Y component. Therefore, actually it is only the X component which get got contacted and not the Y component. So, this is a point which I have mentioned here specifically that we note that only X component of the rods length is contracted and not the Y component. Similarly, not even the Z component because Y prime turns out to be equal to Y Z prime turns out to be equal to Z. Therefore, there is no gamma factor involved gamma factor involved is only in the X component and that is the factor which is the relative velocity direction along which the contraction would really occur. Now, let us take another example. There is an observer in S and he finds that a lightning strikes at a distance of 20 kilometers from origin at t is equal to 0. So, for example, I am sitting here in this particular place and 20 kilometers away from me lightning strikes it could be any other event but I am sort of taking as a lightning. So, at 20 kilometers away some lightning strikes and that was time t is equal to 0 as far as my I am concerned. So, I am sitting at my origin and I find out that at time t is equal to 0 lightning had struck at distance of 20 kilometers away. Remember at this time t is equal to 0 according to me the origin of S was also passing by my side. So, the observer in S would feel that the origin of S prime passing by his side and this event occurring at t is equal to 20 kilometers happened exactly at the same time which is t is equal to 0. Now, the question is at what distance from origin did this event of lightning occur as far as the observer in S prime frame of reference is concerned. This S prime observer is moving relative to me with a speed of 0.6 c. So, let us read the statement again according to an observer in S it lightning strikes at distance of 20 kilometers from origin at t is equal to 0. At what distance from origin did this event occur in S prime which moves with a speed of 0.6 c in frame S. Let me first try to give a wrong answer for this particular example and my experience tells that very large number of student have a tendency of giving this wrong answer. Then we will see what is the correct answer and why the wrong answer is not really correct. So, standard wrong approach to this particular problem is essentially applying blindly the formula of length contraction. So, I may start thinking that at t is equal to 0 and t prime is equal to 0 the origins of S and S prime were coincident. Let us imagine a stationary rod in S frame that in S frame there is a stationary rod which is 20 kilometers long very long rod which is extending from the origin to the place of lightning. I can always imagine that there is a huge long rod which is going right from origin 20 kilometers away and this is stationary in S frame. Now, I expect that the length of this particular rod would be contracted in S because this length of the rod is proper in S frame. So, probably I can apply length contraction formula. Let me just read before I go to the next transparency. At t is equal to 0 and t prime is equal to 0 the origins of S and S prime were coincident. Imagine a stationary rod of 20 kilometers extending from origin to the place of lightning in S. The length of this rod would be contracted in S frame. So, we just now said that gamma is equal to 1.25 for v is equal to 0.60. So, gamma is 1.25. So, this 20 should be divided by 1.25 because this is a contracted length and that answer is 16 kilometers. Therefore, this particular lightning must have struck in S frame at a distance of 16 kilometers from origin. Let me read. So, the coordinate of the lightning in S frame would be given by the contracted length 20 divided by 1.25 which is 16 kilometers. I am again repeating this is the wrong answer. I have written boldly wrong answer. Let us try to find out first correct answer. Correct answer. Whenever there is a confusion apply Lorentz transformation directly. We are not going to make a mistake. Apply Lorentz transformation. I know the coordinate of this particular event which is at x is equal to 20 kilometers, t is equal to 0. I can always find out what is x prime and what is t prime by applying Lorentz transformation. So, this is the correct answer. The coordinate of the event in S is x is equal to 20 kilometers and t is equal to 0. Hence, using Lorentz transformation, the x coordinate in S will be given by x is equal to gamma, x minus vt. x is 20 kilometers, t is 0. So, this particular quantity becomes 0, 1.25 multiplied by 20, 25 kilometers. In fact, this length appears not to be contracted but extended. So, according to the correct answer or according to the answer which you have obtained from Lorentz transformation, this particular event occurred at a distance of 25 kilometers from the origin, not at 16 kilometers, not at 20 kilometers but at distance of 25 kilometers. The length does not, has not contracted. Is it, is it not, does not it look funny? Let us try to find out what is, what is the reason why we are getting different answer. Let us try to understand and let us try to really think that what, why this earlier result was wrong and why this particular result is correct. Why this discrepancy? Look at time. Let us do first a time transformation. t prime is equal to gamma, t which is 0, minus v which is 0.6 here, rate of velocity between the frame, x coordinate of the event which is 20 kilometers, 20 into 10 to the power 3 meters divided by c square. If I take c is equal to 3 into 10 to the power 8 meter per second, I find t prime is equal to minus 5 into 10 to the power minus 5 seconds. What does this negative, see there is a minus sign here. So, this thing becomes minus. What does that minus sign means? It means this event occurred before 0 time. Remember, according to s prime observer also at t prime is equal to 0, his origin was coincident with the origin of s. And this event t prime, this particular event of lightning struck, according to s prime at t prime is equal to minus 5 into 10 to the power minus 5 seconds. It means it appeared before. It means his, the origin of s and s prime were not coincident by till that particular time when the event occurred. The two origins became coincident at a later time, at 5 into 10 to the power minus 5 seconds after this particular event occurred in s prime frame of reference. This is somewhat shocking, but this is what will be the perception s prime frame of reference, observer setting on s prime frame of reference that this particular event occurred before the origins of the two frames were coincident. This is what I have written. So, according to s prime, lightning had struck 5 into 10 to the power minus 5 seconds before the origins of the two frames coincided. In s prime, the events of the two origins coinciding, if we call this as another event, that the two origins being coincident, this event of the two origins coincident and the lightning striking are not simultaneous though in s they are. According to an s observer, lightning striking and the origin of s prime passing both occurred at time t is equal to 0 same time. Therefore, the events were simultaneous. According to t prime, the first the event of lightning struck before the second event that is the coincidence of the origin occurred later. They were not simultaneous. Simultaneous is relative. This is what we have been discussing by giving many examples that this is something which is very different from the classical ideas to understand that simultaneity is not relative, is relative in special theory of relativity. According to s prime, the origin of s is approaching him with a speed of 0.6 c. If we are looking with respect to an observer in s prime frame of reference, this particular origin s was actually moving towards him and only after 5 into 10 to the power minus 5 seconds will the origin s reach his origin. So, this origin is always coming towards him and it will take 5 into 10 to the power minus 5 seconds before this origin will really come and meet him while lightning has struck earlier, while lightning has struck here when the origins were not coincident. This is what I have written. According to s prime, the origin of s is approaching him with a speed of 0.6 c and only 5 into 10 to the power minus 5 seconds after the lightning is striking will the origin of s reach his origin. I wanted to make one particular point which sometimes again I find that students are somewhat confused. See, it is not important when the observer in s prime gets the information of lightning because the information will take certain time to reach. What is important is the time of the event. Suppose one of our friends is departing from railway station at a particular given time and I come to know that his train departed exactly at 10 o'clock. This information may not reach me exactly at 10 o'clock. The person who has gone to see him off may come here, may take 1 hour to reach my place and may give me information that the train had departed actually at 10 o'clock. It does not mean that the event occurred at the time when I got the information, event had occurred earlier. See, what is important is the time of the event. I know that the train departed at 10 o'clock even though I might have got the information late. So similarly, an observer in s prime may have gotten this information later because it always takes finite amount of time for the information to reach him. But all this time if he tries to calculate and take into account, he can find out at what time the event actually takes place. So for example, from the railway station if somebody sends me light signal saying that train has departed, I know how much time the light would have taken to reach me here and I can calculate at what time this particular event would have occurred. So when I am saying the time of the event, it is important to realize that this is the time at which event actually occurred. So this is what I have written. Note when the observer in s prime gets the information of lightning is not important. What is important is the time when the lightning took place as per his calculation. Now as we have said here that after 5 to 10 power minus 5 seconds, the origin of the two frames will reach and at that time according to s prime will be time t prime equal to 0. I can find out what is the distance here. How do I find the distance? I know that after 5 to 10 power minus 5 seconds, this origin will reach him, will reach here and I know what is the speed which is 0.60. So I can calculate what will be this distance at the time when lightning struck. So what can find out what is the distance of the origin of s from the origin of s prime at the instant lightning struck according to s prime observer. Remember all these things I am talking now with respect to s prime observer. We should not never confuse the frame of reference. When I am talking one frame of reference, let us assume that we are sitting in that frame of reference and try to get all the information in that frame only. So I calculate this particular time 500 to the power minus 5 seconds multiplied by the speed with which the origin is approaching me. This distance is 9 kilometers. So let us now recall our ideas, readjust our things, assume that lightning struck at point P and a stationary rod OP of 20 kilometer length is at rest in s as we have said earlier. This length is proper. This length would indeed appear to be contracted in s prime and would indeed be given by 16 kilometers. About that particular aspect there is no problem. The problem is only that this particular length contraction whether my coordinate gives me the correct length or not. This is what I have written here. But according to s prime the lightning occurred before his origin coincided with s. Hence the coordinate of the event does not measure the length of their ordnance frame. While what I had calculated by length contraction is the length, but the question that has been asked is what is the coordinate of that particular event and that coordinate is not equal to the length of the rod. So I have given the situation in a picture. When the lightning strikes, assume that lightning strikes at that instant the picture according to s, the picture that s observer will draw. He will draw that the origin of O prime and O were coincident at that time and at a distance of 20 kilometer this lightning struck. This is the way s is going to picturize the event. But according to s prime the picture will be somewhat different. According to s prime he will feel that he is 9 kilometers away from the origin. This event actually occurred at 25 kilometers but then this rod OP according to him will be 25 minus 9 kilometers which is 16 kilometers which is indeed the contracted length that we had observed. So that is what is the difference of perception. According to this observer s prime he was 9 kilometers away from the origin when the event occurred. So the coordinate of this particular event is actually 25 kilometers what we obtained from Lorentz transformation. Being 9 kilometers away from here 16 kilometers is actually the length of the rod OP which is indeed the contracted length. Before I end let us discuss one more important aspect of this particular Lorentz transformation which is also very popular which is called time dilation. For discussing time dilation we have to define a proper time interval just like we have defined a proper length. Proper time interval we can define between any two events. There could be any two events one event two we have defined many types of events. We will be defining also in the course of lecture many other events. If these two events occur at the same position in a frame of reference then that time interval which is measured in that frame of reference is called a proper time interval. So let us assume that two events appear to be occurring exactly at the same position. So I am sitting here something happens here then again exactly at the same point something else happens here exactly at the same point. So if I take the time difference between these two events that is what is called proper time interval and I measure the time interval between these two events. What I am trying to say is that if anybody else who is moving relative to me measures the time interval between these two events he will find that this time interval is larger that is why it is called time dilation like the eye specialist dilate the eye of the ball becomes bigger. So similarly this particular time interval becomes dilated it has become larger that is why it is called time dilation. In any other frame the time interval between these two events would appear to be larger than the proper time interval that is why it is called time dilation. Now let us assume that it is the S frame we have already discussed what is S frame and S frame that is the S frame in which the time interval is proper. It means these two events must have occurred at the same position in S frame. If they have occurred at the same position it means even their coordinates X coordinates would be same. So it means these two events even one and even two occurred at the same value of X and let us suppose the time as measured by an observer in S is T1 and T2 for these two events. So T1 minus T2 or T2 minus T1 whatever you want to call the difference of the time interval is the proper time interval as measured in S because in S the two events occurred at the same value of X or same position. Now I apply Lorentz transformation if I apply Lorentz transformation exactly like before I get T2 prime is equal to gamma T2 minus V X upon C square this X I have taken to be same. So this X I have not written X2 for the first event I have written T1 prime is equal to gamma T1 minus V X by C square this X being same. I take the difference of these two again these two will cancel out I will get T2 prime minus T1 prime is equal to gamma T2 minus T1. This is what I have written here the time difference between these two events in S is therefore given as T2 prime minus T1 prime which is equal to gamma T2 minus T1. This T2 minus T1 was the proper time interval because this was measured in a frame of reference in which the two events occurred at the same position. So this is proper time interval gamma being greater than 1 T2 prime minus T1 prime will be larger than T2 minus T1 and therefore time interval will be dilated. Many times this proper time interval between the two events is written as tau this is one of the symbol which is very very commonly used. There are one or two small comments which I would like to make first that you will note that to apply time dilation only the X coordinate of the two events had been used. Strictly speaking if the Y coordinates of the two events were different one event occurred here and another event occurred at some other height. Here another event occurring at some other height but at the same position is still time dilation formula is applicable. So strictly speaking that time interval is not proper. We will define exactly correct definition of proper time interval little later but as far as application of time dilation formula is concerned only the X coordinates of the two events need be same. So that is what I have written. We note that to apply time dilation only the X coordinates of the two events have to be same. Though the general definition of proper time interval all the coordinates have to be same. As we said we shall discuss this aspect in detail later. The second point which I want to mention is that if I apply Lorentz transformation for X coordinate according to S prime observer these two events would indeed appear to be occurring at different position. Like in the length contraction according to an observer in S prime the time difference the two events was not same. It was different. Here the X will be different but this is not so shocking because even classically that is what we see. See if any event for example occurs here even classically we do not have to go to relativity. If an event occurs let us suppose somewhere here and one frame was here another frame was let us say also supposed to be here they observe these two events exactly at the same value of X. But the second event when it occurs the second observer has moved to the right therefore the event the X coordinate has changed. Therefore the X coordinate being different is not all that shocking. This is also expected classically. It is a time difference being different in different frame that appears to be much more shocking. So that is where I will end my lecture just by giving small summary. We discussed two important consequences of Lorentz transformation namely the length contraction and time dilation. Then we gave some examples of length contraction and one where one can make an error in the direct use of the formula. Thank you.