 Hello, in our last lecture we had discussed in detail the Michelson-Morley experiment. Michelson-Morley experiment was one experiment which was designed to see if there is something like ether in which earth is actually moving. The motion of earth in ether was considered to be essential if we can believe in the classical postulate that there is a special inertial frame of reference or there is something which we can call as an absolute rest. When we describe the experiment we realize that the outcome of the experiment was negative and it did not seem to show any effect that there is an ether or there is an absolute frame of reference. We also sort of describe that there are some experiments which were also done with totally different nature in mind looking especially at the electromagnetic theory and they were also unsuccessful. So it appears that there is something wrong with the concept of ether which we had sort of discussed in our first lecture without going too much into detail. Now I come to the Einstein's postulates of special theory of relativity. It is not very sure whether Einstein was aware of the results of Michelson-Morley experiment. He was a totally unorthodox thinker. He had firm belief on what his perspective of nature is therefore he made two postulates of special theory of relativity which we will describe now. The first postulate says that laws of physics are same in all inertial frames of reference. No preferential inertial frame exists. It means there is nothing like ether, there is no absolute rest, there is no special inertial frame of reference. All inertial frame of references are equivalent and laws of physics can be applied in one inertial frame as well as in any other inertial frame of reference. Now we have discussed that if we believe purely in the classical theory and purely in Galilean transformation then this will lead us to a value of C which is frame dependent and because C was related to the fundamental constants we have said that this creates a problem. Now in order to make all the inertial frames of reference equivalent C appeared the speed of light appear to be a way of distinguishing between different inertial frames. So here comes the second postulate of Einstein that the speed of light is same in all inertial frames which is a very bold statement saying that irrespective of the frame in which you are you will always measure the same speed it is not C minus U or C minus V or whatever it is. First postulate imply, first postulate implies that there is no absolute rest, no absolute velocity we can never give an absolute, we can never assign an absolute velocity to any object there is nothing like ether, space is not filled with anything, we require something real to which we can attach ourselves and with reference to that frame only we can talk about speeds, maybe it is a dust of particle, maybe whatever it is but there has to be something real on which we can erect our frame of reference, we can erect our axis and only with respect to that particular frame of reference I can talk of velocities, I cannot talk of an absolute velocity, I cannot ask a question what is the speed of sun, what is the speed of earth without specifying which frame of reference I want to measure this speed. So only speeds have meaning when we describe it in terms of a frame of reference. This is what the first postulate of Einstein would mean, the second postulate would mean that irrespective of whether my observer is here, we have said that let us suppose this is a source of light and this is stationary in a frame of reference of this particular observer, he measures the speed of light as C, there is one particular train compartment which is moving towards the source of the light, this particular person would also measure the same speed C, there is one particular person which is in this compartment which is moving in the same direction as the speed of light, he will also measure the speed of light to be C, so there is no C minus U, there is no C minus C plus U, all these three observers will measure the same speed of light, obviously it violates the normal velocity addition formula or relative velocity formula, therefore if we believe in the postulates of Einstein or postulates of special theory of relativity, we must change our ideas, we must change our velocity addition formula. So we require a totally new approach, we are essentially now in dark, we have to really look how to go ahead and try to solve this particular problem, we have just now said that it is not possible to accept the postulates of special theory of relativity in classical framework, because that would give a speed of light that will make speed of light frame dependent, relative velocity formula requires modification in order that it ensures that the speed of light is same in all the frames, so we must change our relative velocity formula. And in addition to that whatever we derive as the new relative velocity formula that must be consistent with the first postulate, it means that must make all the inertial frames of reference equivalent if we believe in whatever we are saying about special theory of relativity. And I would like to add that the measure speeds must be consistent with the velocity addition formula, because eventually the test of any theory, any model lies in whether it connects, explain the experiments, so whatever we are saying must be experimentally viable, it must be seen experimentally, experimentally verifiable, then only we will believe in whatever we are saying as a postulate. Now, let me introduce a concept which is probably a very simple concept and initially I will discuss a concept which is purely in a classical way, so probably everyone should be aware of this particular concept. Only thing because I am going to use some formal terms, so therefore I am describing these things little between more detail than probably necessary, but I think it is essential to get our ideas clear. I introduce the concept of what we call as a transformation. Transformation means very simple thing and I will give you example of what actually transformation is. It effectively means that suppose there is a particle which is moving, let us imagine that this particular particle is moving in this particular direction, there is one particular particle. I know its speed, I know its position at a given time. So, I know let us suppose that this particular particle is at x, y, z coordinates at a given time and I know its velocity. Now, as we have just now said that this velocity is a frame dependent quantity. So, this u when I am talking must be relative to a frame. Similarly, this x, y, z coordinates are relative to a particular frame of reference. Now, if I move, if I change my frame of reference and go to any another inertial frame of reference, I remind you that whenever I am talking of frame of reference is here, we are always talking of inertial frame of frames because we are talking of special theory of relativity. So, unless specifically mentioned a frame of reference we would mean only in inertial frame. So, same particular particle is now being viewed by a different observer which is in a different frame. Obviously, he will measure a different speed, he will measure the coordinates to be different. Now, if there is an equation or there is a set of equations which relate these coordinates, these velocities as observed in one frame to that will be observed in any other frame. These equations, these set of equations are called transformations. So, there can be a coordinate transformation where if I know the coordinate of a particular particle, this coordinate of this same particle in a different frame of reference can be obtained by using these transformation equations. Or it could be a velocity transformation where I know the components of the velocity u x, u y, u z and then I want to find out what are the components of the velocity as observed in a different frame of reference as prime. So, this is what is essentially called a transformation. As I said, it is a very simple concept only thing the name, we are sort of putting a name to formalize it. So, this is what I have written here, the concept of transformation. A set of dynamical variables are known or are measured for a particle in a given frame. So, like for example, in this case we talk about the coordinates, we talked about the velocity. We are interested in finding out the values of the same set of variables. It means if it is coordinate, we are interested in finding out the values of the coordinates. If these are velocity components, then we are interested in finding out the velocity components in a different frame. The equations that we use to change these variables from one frame to another frame, these are called transformation. And as I have just now given you example, this example for example, could be a coordinate transformation or a velocity transformation. So, let us first start with the most simple transformation, which is a classical transformation, which we call as a Galilean transformation. Before we talk about this particular thing, let us define our problem. See as we will soon see, the special theory of relativity can be fairly complex in terms of equations and we do not want to start with most difficult problem right in the beginning. So, without losing generality, we can always define a set of axes which are convenient to us, so that eventual equations that we are going to deal are going to be simple. So, we assume that two sets of frames, S and S prime. This is what I have written here. One is a red frame, which is called S. It has its x-axis, its y-axis and z-axis. So, remember in physics, we are free to choose the set of axes that we want to describe. Physics is not going to change if I choose a different set of axes. Now, there is another frame of reference here, which I am putting in blue in color, which I am calling as S frame of reference, which has also its x-prime, y-prime and z-prime axes. Now, I am making certain conditions on these particular set of axes, as I said, to make our life simple as far as the future equations are concerned. So, if I know the relative velocity of direction between these two frames, I decide to choose my x-axis along that particular direction. It means I assume that S frame of reference is moving in such a way that its origin is always along the x-axis of S frame. So, it always moves along this particular line. So, x and x-prime axes are always coincident with each other and origin is moving just along the x-axis. Similarly, we can of course say that the origin O is moving along the minus x direction of S frame of reference, because if there is an observer in S frame of reference, he would feel as if this particular frame of reference is going behind him, because observer sitting its own frame does not see the motion of its own frame. Like when we are sitting on earth, we do not notice the motion of earth. We feel the earth is stationary and we accordingly change our idea of directions of motions of let us say sun or moon. So, similarly, if a person was sitting on S frame, he would notice that this O is going behind him along minus x-prime directions. So, this is the direction of the relative velocity between the frames. I assume that y-prime axes is always parallel to y-axis. I assume that z-prime axes are always parallel to z-axis. So, this is the special set of axes that we have chosen to describe special theory of relativity. It actually does not lose generality, but because we can always know the relative velocity direction and along that particular direction, I can choose my x-axis. As I have said, physics is not going to change if we choose a different set of axes. So, this is a special set of axes which we have chosen. So, this is what I am describing in my next transparency that these are the conditions that I have put on the two frames. The frame S-prime moves with respect to S along x-axis with a velocity v. Let me just say that the velocity symbol v is reserved for the relative velocity between the frames. Then the initial condition that we are putting that clock, see when there is a relative motion and O-prime is moving along the x-axis of O, there will be some time when O and O-prime will be coincident. Let me go back to my old transparency. Here, because there is a relative motion, so there will be a time when O-prime and O will always be moving, will always be coincident when O and O-prime will always be coincident. We assume that that is the time when the watch of S and S-prime both are set equal to 0. So, at that time S also makes the time t equal to 0. S-prime observer also makes time t prime to equal to 0 in his frame. Let me also reassert one of the things because the frames of reference that I am going to talk are inertial. So, this velocity v that I am noticing writing here is constant. It does not change because if v was changing as a function of time, then at least one of the frames would be non-inertial. So, it means essentially that v has to be constant because the two frames S and S-prime that I am talking are both inertial. Now, let me come to Galilean transformation. As I say, it is comparatively simple transformation and purely classical transformation. So, this was my S-prime frame. Let us assume that we are talking of one particular particle here and the observer S measures the coordinate of this particular particle in his frame and finds out to be x, y, z. My question is that what will be the coordinate of the same particle as seen in S-prime frame of reference? So, observer sitting in S-prime, what will be the coordinates of this particular particle as measured by him? As we can see because this was your y-axis, this was your z-axis because y and y-prime axes are always parallel to each other. So, this particular component y-component will always be same. This y-component will always be same whether S you are talking of S or S-prime frame of reference. Similarly, z-component will also be same. So, the coordinates y and z will be same as measured in S and S-prime. But it is x-coordinate which will be different because according to the observer in S-prime, he would measure this as the x-coordinate while according to the observer in S, the x-coordinate will be this. These two x-coordinates will be differing by this much amount which is essentially the separation of origins between O and O-prime. So, this coordinate x as measured in S will be larger from what was measured by S-prime by this much amount. And this amount, this distance O and O-prime would depend on how much time has passed because the relative velocity is v. So, if this particular measurement was done at time t, then this particular origin would have displaced by a distance vt. Remember time t was 0 when the two origins were considered. So, this distance will be vt. Hence, I expect that x-prime would be equal to x-vt. y-prime would be equal to y and z-prime will be equal to z. This is what is called Galilean transformation. I will take a minute of time to explain this particular thing. We have also written clearly in this equation that t-prime is equal to t. Time is something which normally we have not been talking in classical mechanics. We have just been silent. We just sort of ignore. We assume that there is a uniform flow of time in all the frame of references. So, once the two observers in S and S-prime have measured their time, whenever a measurement of the coordinates of the particle is done, whatever was the time in S-watch, S-watch, same will be the time in S-prime's watch. So, we assume that t and t-prime is equal to is same. Even though classical mechanics, we have not discussed it. We have never talked specifically about the time, but it is a sort of implicit assumption that the two observers, once they have synchronized their watches, then their times will always be same whenever we make a measurement or whenever an event occurs. So, we assume specifically that t-prime is always equal to t. Let me call a transformation as inverse transformation. Actually, inverse transformation is nothing but same as direct transformation, except that we are changing frames from S to S-prime. It means the information is available or the coordinates are available in S-prime frame of reference and I want to find out the coordinates in S-prime of reference. It means I know what is X-prime, Y-prime, Z-prime and I want to find out what is X, Y is Z. So, it is just inversing because there has to be some sort of symmetry. So, it essentially means that all you have to do is to change V to minus V because the equations have to be same. So, what we call as inverse transformation is if I know X-prime, then I will find out X as X-prime plus V-t-prime, Y is equal to Y-prime as we have discussed, Z is equal to Z-prime and of course, it assumes implicitly that t is equal to t-prime. The two times are same. So, this is what was the Galilean transformation which involves the transformation of the coordinates of a particle. Now, let us look at the velocity transformation. As we know very from the standard traditional classical mechanics that the speed of a particle is given as dx dt. I must write u x because I am differentiating only x. Similarly, u y will be given as dy dt, u z will be given as dz dt. Let me just spend a couple of minutes on my notations. I am using the symbol u here to define the particle velocity. See earlier, remember we had used a symbol v. v was the relative velocity between the frames. The symbol u I have reserved or it is generally traditionally reserved for the particle velocity. Remember the particle need not be moving with constant velocity because particle could be accelerating. It is only v which is constant as a function of time because this is the relative velocity between the frames. But both the observers s and s-prime could be observing some other particle and that particular particle could be accelerating, could be stationary, could be moving with any velocity. So, what we are talking is the instantaneous velocity of a given particle. So, the instantaneous velocity of a given particle is given in terms of the derivatives dx dt dy dt dz dt. And using Galilean transformation which we have just now used which says that x-prime is equal to x minus vt. We can very easily write, we have said x-prime is equal to x minus vt. If we differentiate with respect to time, the speed as measured in prime frame of reference would be given by dx-prime divided by dt-prime. Remember, I must differentiate with respect to the time measured in the observer s own frame of reference. But at this moment we have said t is equal to t-prime so we need not bother about it. We differentiate this with time I get dx dt minus v. This is what we have written in this particular transparency dx-prime dt-prime is equal to dx dt minus v. Similarly, because in Galilean transformation y-prime was equal to y, therefore dy-prime dt-prime is equal to dy dt. Similarly, dz-prime dt-prime is equal to dz dt because this is the velocity, x-component of the velocity measured in s-prime frame of reference. This is the x-component of the velocity measured in s-frame of reference. I can write the equation as ux-prime is equal to ux minus v. Similarly, this equation can be written as ui-prime is equal to ui. This equation can be written as uz-prime is equal to uz. So, this gives me my velocity transformation equation which means that if I know ux, ui, uz, I know how to find out ux-prime, ui-prime, uz-prime. Actually, this transformation is as simple as the relative velocity, standard relative velocity because relative velocity as you know is given by this same expression. So, velocity transformation is same as the relative velocity expression with which we are quite used to it. Similarly, I can write a inverse velocity transformation. It means if the velocity components are known in s-prime frame of reference, how do I find out the velocity components in s-frame of reference? And as we have discussed, the prescription is very easy. Just change v to minus v. So, it becomes ux-prime plus v, ui becomes equal to ui-prime, uz becomes equal to uz-prime. This is what is called inverse velocity transformation. Now, let us take few examples of Galilean transformation to illustrate my point specifically to initiate little bit of discussion on the concept of t is equal to t-prime. So, let us assume that there is one particular particle which is moving with a speed u or with a velocity u which is going in this particular direction, some arbitrary direction which is being observed by a frame s because I have not put a prime u on u. It means this is the velocity as has been observed in s-frame of reference. Now, my question about this particular problem is given in the next transparency, which says that an observer in frame s locates a particle at origin at t is equal to 0. So, at t is equal to 0, this particular observer feels that this particular particle was at the origin. And this is moving with a constant velocity of 5 meters per second making an angle of tan inverse 3 by 4 with x-axis. I have taken some arbitrary velocity, I have taken some arbitrary angle just to illustrate my point. So, this particular particle which was moving here is speed is measured by observer s and is found to be 5 meter per second. And the angle that this makes, this speed makes with s is given by tan inverse 3 by 4. This is what I have, just assume. Question is that find the position of the particle in s at t is equal to 2 seconds. It is a very simple problem, a typical standard classical mechanics problem. What is the position of the particle in s-frame of reference at t is equal to 2 seconds? Then you assume that there is another frame s, the one which I have drawn as blue lines in my earlier figure. I assume that this observer, this frame s-frame is moving with a speed of 1 meter per second. I mean some arbitrary value which I have taken. Assume that the two frames obey the condition of Galilean transformation described earlier. It means I assume that y-axis is parallel to y-prime axis, z-axis is parallel to z-prime axis. Relative motion is only along the x-direction and the origin of s-prime always moves along the x-axis of the origin of s. And also the two times are measured when the two frames, the two origins were coincident. So, these were the conditions under which we have derived Galilean transformation. So, we assume that these conditions are also true for these two frames s and s-prime. Now, the question is that what is the speed of the particle and its coordinate at t is equal to 2 seconds also in s-prime frame of frames. So, we have to not only find the coordinates in s-frame, we have also to find out the coordinates in s-prime frame. So, this is what is my question. First, we find out the coordinates of this particular particle in s-frame which is very simple because I know the velocity. So, I must be able to find out what is the x component of the velocity and the y component of the velocity. We have said that tan theta is equal to 3 by 4. And we know that tan theta can be written as sin theta divided by cos theta. Also, we know that sin square theta plus cos square theta must be equal to 1. So, if you solve these equations, you will get that sin theta will be equal to 3 by 5 and cos theta will be equal to 4 by 5. As you can see, tan theta is sin theta divided by cos theta. So, if I divide these two, I get 3 by 4. And if you square and square this and add, this will give you 5 by 5 which is equal to 1. It will be 5 square divided by 5 square which will eventually give you 1. So, sin square theta plus cos square theta will actually turn out to be equal to 1. So, the value of sin theta is 3 by 5 and cos theta is 4 by 5. So, when I write ux, I know that this is given by u cos theta. Similarly, ui is given by u sin theta. We have just now said sin theta is 3 by 5. So, ui will be equal to 5 multiplied by 3 by 5. We have said cos theta is 4 by 5. So, ux will be given by 5 multiplied by 4 by 5. Therefore, the x component of the velocity of the particle is 4 meters per second, while the y component of the velocity of this particular particle is 3 meters per second. And of course, there is no speed in the z direction under the assumption, under what has been given in the problem. So, this speed is 0. If you have to find out the coordinates, all we have to do is to use the simple kinematic equation. I have assumed that this particle is not accelerating. So, acceleration is 0. So, the displacement is just given by ux multiplied by t or ui multiplied by t as the case may be. So, along the x direction, the x component is given by ux multiplied by t. ux we have just now seen is 4 time, which has been given as 2 seconds. This turns out to be equal to 8 meters. Similarly, y will be ui multiplied by t. ui we have just now calculated is 3. 3 multiplied by 2 is 6 meters. So, eventually the coordinate of the particle at t is equal to 2 second in S frame will be given by 8 meters and 6 meter. x component is 8 meter, y component is 6 meter. Now, if I want to find out the coordinate of this particular particle at the same time in S frame of reference, all I have to do is to use a Galilean transformation. So, this is what I have used in this particular transparency. So, x prime is given as x minus vt. This is what we have just now calculated. x was equal to 8, which we have just now evaluated. v is the relative velocity between the frame, which has been given as 1 meter per second. 2 is the time. So, we get x prime is equal to 8 minus 1 multiplied by 2. This being equal to 2, x prime is equal to 6 meter. According to Galilean transformation, y coordinate does not change as we change from S to S prime frame of reference under the conditions that have been described. Therefore, y coordinate remains exactly same, which is 6 meter. Of course, Z coordinate also remains same. Z coordinate initially was 0. It remains 0. It continues to remain 0 in S prime frame also. Now, if I want to find out the velocity of the same particle as S frame of reference, I have two methods. One is that I realize that the time interval measured between 0 and t. See, particle started, let us say, at t is equal to 0. And eventually, the next measurement of the particle's position was done at t is equal to 2 seconds. So, this time interval of 2 seconds remains identical in S and S prime frame of reference. It means, when the observer S measured the particle coordinate and he found time at that particular time in his watch as 2 seconds, according to the observer in S prime frame of reference, his watch also showed a time t is equal to 2 seconds. Therefore, whatever coordinates have changed, whatever coordinates he has measured, according to him, the particle, remember, at t is equal to 0, the origins of both the frames S and S prime were coincident. So, according to observer in S prime, this particular particle has moved along x direction, not by 8 meters, but by 6 meters. Because it started at X prime is equal to 0, but only landed up at X is equal to X prime is equal to 6 meters. And that particle took the same amount of time as observer in S frame has evaluated, which is 2 seconds. So, he would calculate U X prime and find it out to be 6 divided by 2, which is 3 meters per second. And U Y prime, of course, will be Y coordinate as measured by his frame, which is the same as the Y coordinate measured in X frame, which is 6 meters divided by time measured in his frame, which is same as the time measured in S frame. Therefore, U Y prime is equal to 6 divided by 2, which is 3 meters per second. This is very, very simple ideas, but again I am less sort of emphasizing on these things, so that the comparison with relativity becomes easier. As I have said that we could have also used a velocity transformation formula without using the time concept. And then we would have got U X prime is equal to U X minus V, U X we have just now calculated as 4 meters per second, V is 1 meter per second, so U X prime will be 3 meters per second, U I prime is going to be same as U I, which is 3 meters per second. So, we get the same answer from the two approaches that the X component of the velocity as measured in S frame of reference will be 3 meters per second, Y component of velocity as measured in S frame of reference will be 3 meters per second. Obviously, the components have become different, more specifically the X component of the velocity has changed once I have changed the frame of reference. This is something which is very, very normal, which we have not been discussing right from our first lecture that the speeds are frame dependent quantities. If I am interested in finding out the speed of the particle in S frame of reference, I have to take overall the mod of all these velocities, which means the velocity would be given by U prime is equal to under root U X prime square plus U I prime square plus U Z prime square. This we have calculated as 3, this we have calculated as 3 meters per second, so this becomes 9 plus 9, eventually if you evaluate U prime, the speed will be 3 multiplied by root 2 meters per second. So, the speed which was observed in S frame was 5 meters per second, the speed of the same particle as measured in S frame of reference turns out to be 3 multiplied by root 2 meters per second. Now, let me take another example, which is a more interesting example and it discusses a concept which again we will be discussing quite a bit in relativity, which is the concept of simultaneity. We call two events to be simultaneous. See, let me say at this moment that relativity many times we talk of events. See, for example, in the example which we have given earlier, when observer in S measure the coordinate of the particle, whatever might be the coordinates, whatever might be the time, we will call that as an event. So, a particular observer measures the position of the particle, that is an event. This same event, the measurement of the coordinate of the particle is being observed by another frame of reference, another observer in a different frame of reference. So, an event is an event, which is observed in all the frames of reference, only thing which could be different are their dynamical variables. Their coordinates could be different, their velocity components could be different, but what we are measuring are the coordinates or dynamical variables corresponding to a given event. So, if these two, there are any two events which occur at the same time, we call them a simultaneous events. If we are seeing in an observer and finding out two events, for example, if we are sitting here on the earth and we find that a plane taking off and let us say, train departing and we find that both these events occur at the same time. They may occur in a different position, the plane may take off from airport, while train may depart from a particular railway station, but if they happen according to me, in the same time, I will call that these two events were simultaneous. Now, in classical mechanics, if two events are simultaneous in a given frame, they remain simultaneous in all other inertial frames. So, this is an example which I am giving, again using Galilean transformation, describing the fact that two events, if they happen to be simultaneous in one frame, will turn out to be simultaneous in another frame. Of course, this is one specific example, but this is a general statement. So, my example is as follows. An observer is sitting or standing, whatever you want to call it, exactly halfway in a running compartment of length L. So, there is a train compartment which has a total length of L and he is sitting or standing just exactly half the way within the compartment. He throws two walls at the same time, which I am assuming at the time t prime equal to 0 with speed u prime as measured by him. Again, I remind you I am using symbol u because this is an object whose velocity is being measured in S and S frame of it is two different frames of, in this one frame is the compartment frame. So, in the compartment frame it is found that this particular ball is thrown with speed u prime as measured by him, of course. One towards the front ball, ball and the second one towards the back ball. So, if the compartment is moving, there is a ball which is moving along the direction of the speed, which is the front. So, let us suppose he is standing in between, so that his left hand is towards the motion and right hand is against the motion, then he takes ball in left hand and throws in that particular direction towards the motion and other ball throws in a direction opposite to the motion. Find the time when the balls hit the ball in the compartment and the ground frame. So, these balls go and hit the walls of the compartment one along the direction, another opposite to the direction of the velocity of the compartment and what I have to find are the times when these balls hit the balls. This has been picturized somewhat in this particular cartoon that this is in a compartment which I assume is moving with the velocity v relative to observer on the ground. I assume that ground frame is an inertial frame. This person is sitting halfway exactly in this particular compartment. He has one ball which is pointing towards the direction of the motion. There is another ball which is pointing in a direction opposite to the motion. He throws one ball in this direction. He throws another ball in this direction. This ball goes and hits this wall. This ball goes and hits this wall. I have to find out the time when this happens as far as this observer is concerned assuming that these balls are thrown at time t prime equal to 0. Then I would find the same motion. I will describe the same motion with reference to an observer s, which is standing on the ground. From now onwards, we will try to define the events to make things comparatively simple. So, I describe two events. The first event is the first ball reaching the front wall. It hits and the front wall, the wall which is towards the direction of the motion. So, this is the ball which comes and hits here. This is my event number one. This ball goes and hits the back wall. This is my event number two. So, event number one, this ball coming and hitting here. This event number two, this ball coming and hitting there. These are the two events that I am describing. Event one, the first ball reaches the front wall. Event number two, the second ball reaches the back wall. We are doing all this calculation in s prime frame of reference. We have not, that is the compartment frame of reference. I have not really talked about the ground frame of reference so far. Now, if I want to find out what is the time of event one, that is very simple. This ball has to travel a horizontal distance of l by 2 because the length of the compartment was l. This is my compartment. This length was l. Ball is thrown exactly halfway. So, one ball goes like this, another ball goes like this. The distance this ball, the horizontal distance that this ball travels will be l by 2. Of course, this may have a slight downward displacement because of gravity, but that does not bother about this particular time. That does not change this time. Similarly, this ball also has to travel a distance of l by 2. I am talking everything in terms of this frame of reference, which is the compartment frame of reference. So, this distance travelled is l by 2. This distance is travelled by l by 2. We have also assumed that the two speeds are same. This speed is u prime. This speed is also u prime. So, obviously, the time that this particular ball will take to hit here, looking only at the x component will be this time t will be equal to l by 2 u prime. So, this is what I have written here in this particular transparency that time t 1 prime, this 1 refers to the event 1. This prime refers to the fact that this particular time is being measured in s frame of reference. So, t 1 prime is equal to l divided by 2, which is the length divided by the speed of the ball, which is u prime. So, this is the time t 1 prime. Similarly, time t 2 prime referring to event 2 is given by t 2 prime is equal to l divided by 2 u prime. So, these events are simultaneous because t 1 prime is equal to t 2 prime. So, the observer in s frame of reference will conclude that these two events, the ball hitting the front wall and the ball hitting the back wall are occurring at the same time and therefore, they are simultaneous events. Now, let us look at the same situation as observed with respect to the ground frame. So, there is an observer sitting here at the ground. Let us assume that at the moment when the balls were thrown, the origin of this particular person was also coincident with the origin of s prime. It means these two observers were just sitting just side by side. Therefore, when s prime, s prime's watch showed t prime equal to 0, the watch of this particular observer also showed time t is equal to 0. However, because this particular train is moving with a speed v in the frame of the ground observer, his perception about the speeds and distances would be very different. Why it would be different? Because this ball, which has been thrown in this particular direction, according to this observer, he would find that the speed of the ball is different using standard relative velocity or velocity transformation formula. So, this speed of the train will get added up. So, in fact, what he would find that the speed of this particular ball is larger than u prime. Similarly, this particular ball, which has been thrown backwards according to this observer, this ball would not actually be thrown with the speed u prime, because this particular train from which this particular ball was thrown was itself moving. So, he would find that this speed is smaller than this speed of the front ball. So, if we call this as ball 1 and we call this as ball 2, according to the observer on the ground, ball 1 moves with a larger speed than ball 2. But as the ball keeps on moving in this particular direction and tries to approach this particular ball, this ball according to him will also is moving towards the right. So, according to this observer, this ball has to now travel a larger distance, because by the time this ball starts from here and reaches this ball, this ball has already moved ahead, because during this particular time there has been motion. So, this ball may be somewhere here. So, this particular ball has to move a larger distance. On the other hand, this particular ball because it is moving towards this particular in this particular direction. So, by the time it hits this ball, this particular ball would have moved towards it. So, this ball has to travel a smaller distance to hit the back wall, then this ball, ball number 1, which has to travel a much larger distance to hit the ball. So, the perception of the ground observer will be quite different. According to him, the ball number 1 will be thrown with a larger speed, but it has to travel a larger distance to reach the ball. While the second ball has been thrown with a smaller speed, but also it has to travel a smaller distance to reach this particular ball. And what we are going to show just now is that, because this ball has to travel a larger distance with large speed, by making calculation we will show that still the times measured by this particular observer for the two events will turn out to be same exactly as what has been observed by S prime observer. So, let us try to look into this particular calculation, a simple calculation. So, we look at the same two events in S frame of reference, which is the ground frame of reference. So, I calculate the speed of the two balls in the ground frame of reference using velocity transformation. So, speed of the first ball using inverse velocity transformation, in fact, I have what I have used is inverse velocity transformation, because the speeds have been given in S frame of reference and I want the speed in to be S frame of reference. So, remember in inverse velocity transformation it was u x is equal to u x prime plus v, u x prime is equal to u for the first ball and v is the relative velocity. It means it is the velocity of the train as measured by the ground observer. So, the x component of the velocity of the ball as measured by the ground observer will be u plus v. Similarly, the speed of the second ball we apply the same velocity transformation, inverse velocity transformation will be given by same expression u x is equal to u x prime plus v, but for the second ball because it moves in minus x direction, this will be minus u plus v. So, as we have said that this velocity, this x component of the velocity is going to be smaller than this component. We will find that u x for the second ball will be smaller than the u x for the first ball. So, this is what I have written here in this transparency that one can see that the speed of the first ball is larger than the second one, but that ball also travels a larger distance to reach the ball. This is because of the motion of the ball. Now, what will be the observation of the observer in S frame of reference? According to the observer, let us look at the first ball which was the ball which we have thrown in this particular direction. As we have mentioned that this particular ball has to travel a larger distance not only L by 2, but also this additional distance. This additional distance is the distance moved by the train during the time that this particular ball took to hit the wall. That particular time, if it happens to be T1, then the total distance, this distance will be v multiplied by T1. So, the total distance that this ball has to travel is what is written in this particular transparency is v T1 plus L by 2. This particular distance is covered by the first ball with a speed of u prime plus v. So, u prime plus v multiplied by T1 is given by v T1 plus L by 2. If we look at the second ball, the distance travelled now gets reduced because by an amount v multiplied by T2, if T2 is the time when the second event occurs, this will be given by v T2 minus L by 2 because remember there is a negative sign. The motion of the ball is in minus x direction and it travels with a speed of v minus u prime. So, this travels a smaller distance with a smaller velocity. If we just look at these two equations, here v T1 will cancel out with this v T1, here v T2 will cancel out with this v T2. We will get the same expression T1 is equal to L divided by 2 u prime, T2 is equal to L divided by 2 u prime, the same expression that we have obtained from the observation of the observer in S. So, observer in S also finds that the two times are same and he will also notice that these two events are simultaneous. This is what is the point which I was trying to bring home saying that the two events which are simultaneous in S frame also appear to be simultaneous in S frame. This is what I have written, we just see that in ground frame also the time is same and the events are simultaneous. Two statements that the two times are same and the events are also simultaneous. Now, before we go to the next lecture, let me just pose one question, just pose one particular issue saying that eventually Einstein realized that it is this particular time which we are assuming to be same in all the frames. This particular simultaneity which we are assuming to be same in all the frame is something to be suspected. If we are looking into a different set of transformation which is consistent with the postulates of special theory of relativity, probably we have to relook at our concept of time. So, this is what I have written. Next we shall see that the two, in fact, we will work out the same two examples and we will see that if we assume time to be same, then the postulates of special theory of relativity are sort of violated. In order that we have to maintain the postulates of special theory of relativity, probably we will have to have to relook at the time t, something which we are taking for granted so long. I will just give a summary of whatever we have discussed today. We describe the postulates of special theory of relativity, we describe the concept of transformation, we described the Galilean transformation and the velocity transformation, gave some examples for that and then finally we raised an issue over time saying that maybe it is the time which we must suspect. Thank you.