 In our last lecture, we had discussed the various postulates of special theory of relativity. They were two postulates, we will sort of revise them. We also discussed about Galilean transformation, we discussed what is the transformation and how the classical Galilean transformation looked like, we formalize that transformation. We discussed some examples, especially with the idea of time in mind. We saw that in the classical mechanics, time is always treated to be same in all the frames. We also saw that if there are two events which appear to be simultaneous in one frame, it means they occur at the same time. Then another frame also they will appear to be simultaneous. This is ensured by the fact that time is same in both the frames of reference. So, this is what we had discussed in a brief nutshell that postulates of special theory of relativity, Galilean transformation and the concept of time and took some examples to show that its equality in different frames assures that simultaneous events in a frame are also simultaneous in another frame. Just to recapitulate, let me just tell again the postulates of special theory of relativity. First is the laws of physics are same in all inertial frames of reference, no preferred inertial frame exist. We have always said that all inertial frames, the laws of physics should always remain same. You cannot have something like ether, you cannot have something like absolute rest, you cannot talk of absolute velocities, all you have to talk is only relative in terms of a given frame. You must have something really physical to which you must attach your frame of reference and only with reference to that you can talk about speeds velocities. Then the second postulate was that the speed of light c is same in all the inertial frame. I must add whenever I am saying speed of light, it means the speed of light in vacuum. Of course, when we talk of medium, the light of speed of light may get reduced, but when we are talking of the speed of light, c is always the speed of light in vacuum. Again I recapitulate Galilean transformation. This is what we call as a direct transformation. It means if we know the coordinates of an event which happen at x, y and z and it happens at a time t, then the coordinate of the same event as appearing to another frame of reference s is given by this equation x is equal to x minus vt, y is equal to y, z is equal to z and of course, we have said that implicitly we have assumed that t is equal to t. Similarly, we talk of inverse transformation which says that if you know the coordinates of an event in s frame of reference which is x, y and z and it occurs at a time t, then the coordinate of the same event in s frame would appear to be as x is equal to x prime plus vt prime, y is equal to y prime, z is equal to z prime, t is equal to t prime. Of course, this transformation required a special set of axes about which we had discussed earlier that we assume that y and y prime axes are always parallel, z and z prime axes are always parallel. The relative motion is only along the x direction and the origin of s prime moves along the x axis. Time is measured from the time when the origins of the two frames are coincident. This is what we have discussed. Inverse transformation can always be obtained from direct transformation by putting v is equal to minus v. We have also discussed the velocity transformation. It means I know the velocity components in s frame, ux, uy, z. We can find out what will be the velocity components in s frame of reference which turns out to be ux prime, uy prime, uz prime and similarly we discussed about the inverse velocity transformation. We also said that this is exactly same as the normal relative velocity concept with which we are very familiar in classical mechanics. Today I will try to give some example and try to say that how Galilean transformation and second postulate of special theory of relativity do not go along with each other. It means it is not consistent with the second postulate of special theory of relativity. We have mentioned about this in passing by in our first lecture. When we are trying to introduce a special theory of relativity, we had mentioned that from the normal traditional relative velocity, we will never find the speed of light to be constant. But once we have formalized the Galilean transformation, let us try to look with the eyes of the Galilean transformation, try to see how this particular transformation is inconsistent with second postulate of special theory of relativity. We will also give an example where we will see that if second postulate of special theory of relativity is correct, then simultaneity is also relative. It means in a frame if the two events appear to be simultaneous in another frame, they may not. These are the two examples. Let us start with them before we really actually come to a transformation which is consistent with the second postulate of special theory of relativity. So, this is what I have written here. We now show that Galilean transformation is not consistent with second postulate. We also show that simultaneity of events is also relative under the second postulate. Let us go back to the old example which we have discussed in our last lecture. See what we have done in that particular lecture that we assume that a ball is thrown making a particular angle in the xy plane in S frame. We try to find out the coordinate of that particular ball at time t is equal to 2 seconds in that particular frame. Then we found out what are the coordinates of the same event that is finding the ball at t is equal to 2 seconds in S in S frame in S frame and then eventually found out what are the velocity components. So, this is the other problem which we had discussed. Now, I have modified that particular problem. Instead of ball being thrown, let us imagine that we throw a pulse of light. Most of the things remain same other than we have changed some numbers to make it look really somewhat more realistic. Otherwise, the problem is more or less identical of what we did for that particular ball-ball problem in last lecture. So, here we have a pulse of light. Let us assume that this pulse of light is highly localized. So, you can determine to a great accuracy what is the position of the pulse. So, we throw a pulse of light from the origin at the same time when this O prime origin was also coincident which happens to be t is equal to 0 which also gives t prime to be equal to 0. Most of the things as I said have remained same. So, we assume that this is thrown in xy plane and it makes an angle of tan inverse 3 by 4 with x axis. This angle remains same as we have done in the last problem. What we have said now that find the position of the pulse in s at t is equal to 2 into 10 power minus 6 seconds that is 2 microsecond to make numbers somewhat more realistic. Instead of 2 second, I have made it 2 microsecond because the speed of light is very large. And of course, I am assuming that the pulse is highly localized. So, I can really determine its position to a reasonable amount of accuracy. The question is essentially identical. Assume there is another observer in s prime frame which is moving relative to s with a speed of 0.6 c. I have changed this relative speed also to make it somewhat more realistic from the point of view of special theory relativity. Assume that the two frames over the condition of Galilean transformation which we had described just now again. Find the speed of the light pulse and its coordinates at t is equal to 2 into 10 power minus 6 seconds in s frame of reference under Galilean transformation. So, we assume that a Galilean transformation is still valid and try to find out its coordinates and also the speed in s frame of reference. As you can see that this problem is very identical to the problem that we had just now discussed in our last lecture about the ball. The method that we are going to adopt is essentially identical. We find the x component of the speed and the y component of the speed. We have discussed last time that if tan theta is 3 by 4, then sin theta will be 3 by 5 and cos theta will be 4 by 5. Therefore, we know that u x is equal to u cos theta and u y is equal to u sin theta, standard way of taking resolving a vector along x and y direction. So, we will get u x to be equal to 0.8 c, u y is equal to 0.6 c and because the light is been thrown in x y direction, u z is equal to 0. So, this remains essentially identical of what we have done earlier except that the speed is now c and not a classical speed. We can find out the coordinates at t is equal to 2 into 10 power minus 6 second that is 2 micro second again by a simple application of formula that the x coordinate will be u x multiplied by t, which is 0.8 c is the value of u x time is 2 into 10 power minus 6 second, we multiply by the 2 and we get 480 meters. Similarly, y will be equal to u y multiplied by t, we have evaluated u y which turns out to be 0.6 c and t is 2 into 10 power minus 6 seconds. So, y component turns out to be 360 meters and of course, z component is 0. Of course, in this calculation I have taken the speed of light as 3 into 10 power 8 meters per second to make our problem simple. So, we see that the coordinate of this particular light pulse at 2 micro second will be 480 meters, 360 meters and 0. Question is that what will be the coordinate of this particular light pulse as being viewed in S frame of reference assuming Galilean transformation. We apply standard Galilean transformation the equation we have just now given x will be equal to x minus v t, x we have just now calculated as 480, v which is the rate of velocity between the frames is 0.6 c time t is 2 micro seconds, we put it here I will get 120 meters y prime remains same which is equal to y which is equal to 360 meters. So, according to observer in S frame of reference the coordinate of the same event will be 120 meters, 360 meters and 0. So, the 2 observers will find out that the coordinates are different in their frames 1 is 480, 360, 0 another is 120, 360, 0 standard thing expected in a classical mechanics. Now, if I want to find out the speed like we did in the case of wall as seen in S frame of reference this displacement I must divide by the time and in the classical mechanics if you remember we always calculated time to assume time to be same. So, we take 120 divided by 2 micro second which is the same time as seen in S frame and we will find out u x prime to be 0.6 into 10 to power 8 meter per second u i prime will be 360 divided by 2 to 10 power minus 8 which is 1.8 into 10 to power 8 meters per second as you can see that this u x prime has changed. In fact, it has become smaller than u x obviously, we would expect that the speed of light in this particular frame will be reduced if Galilei transformation was correct. I have calculated the speed of the pulse in S prime which is given by standard formula of finding out the length of a vector under root of u x prime square plus u y prime square plus u z prime square which turns out to be approximately 1.9 into 10 to power 8 meter per second while in S the same speed was 3 into 10 to power 8 meter per second. So, as you can see that if Galilean transformation was correct then the speed of light will turn out to be different in different frame as was will turn out to be different in different frames and this is something which was against the second postulate of spatial theory of relating. So, this is what I have written we either see that the speed of light is different in S prime violating second postulate. If we have to find out a transformation in which the speed of light is maintained as C even in S frame S frame frame of reference then either this x prime or t prime or both x prime and t prime have to be changed from what has to be obtained from Galilean transformation. So, we must have a transformation in which still x prime divided by t prime and correspondingly calculating y u y prime eventually leads us to a velocity which is equal to speed of light. So, probably we require a change in x coordinate as well as in type coordinate. Let us look at the second example which we had done in the last time which was an example in which a particular observer sitting in a train throws two balls one towards the motion of the train another against the motion of the train. This particular observer concludes that the two events even number one the ball hitting the front wall and even number two ball hitting the back wall occurs at the same time. We also concluded that the ground observer would also notice that the times are same and according to the ground observer also these two events will appear to be simultaneous it means appearing at the same time. Now, again we replace this particular experiment this particular example and change the ball to a light a pulse of light most of the problem remains exactly identical. So, I have said that an observer is exactly half way in a running compartment of length l prime as we will eventually be seeing that the length also becomes frame dependent. So, because this s prime frame of reference I have decided to call it l prime. So, the length as measured in s prime frame of reference is l prime and of course, in the classical mechanics Galilean transformation will assume that this length is same as seen in by the ground observer also. He shines light instead of throwing balls at t prime is equal to 0 which travels both in front and the back direction. Front direction means the direction of the motion of the train and the back direction is opposite to that. We define events exactly like before even number one light reaches the front wall even number two light reaches the back wall. I put this this particular figure that this observer is sitting in the train which is moving with respect to the ground observer with a speed v. This person has one light source here one light source here which shines light or could have a single line light source does not make a difference shines light which goes this way which is opposite to the direction of the motion which I call as a back wall and a light going towards this direction which I am calling as a front wall. So, a light moving to the right a light moving to the left this is towards the front wall this is towards the back wall and our events are even number one this particular light pulse reaching here even number two this light pulse reaching here at the back wall. Both these events are being observed by another observer s on the ground which we assume to be inertial. Let us look at the motion both from point of view s prime and s when we have a light source rather than two walls being thrown. Let us go to the frame of s prime if you go to the s prime frame of reference this was my train light source was let us assume a single light source this goes in this particular direction this particular light goes in this particular direction. So, each light pulse actually travels a distance of l by 2 according to the observer in s prime the speed of light is c. So, the time taken for each of the event is l prime by 2 c let me replace l by l prime as we have just now said. Now, the distance travelled by both the pulses this and this are same which is l prime by 2 both of them travel with the same speed c. So, the time taken will be l prime divided by 2 c now this is what I have written here both events are simultaneous in s prime that is they occur at the same time and the time is given by t prime is equal to l prime divided by 2 c. Of course, it assumes that the instant when the light pulses were thrown or when the light was we shine the light at that time time t prime was equal to 0. So, the observer in s prime frame of reference would feel that these two events are simultaneous. Now, let us look with respect to an observer in s frame if the observer is sitting here in s prime s frame in the ground and if the second postulate of special theory of light a special theory of relativity is correct then according to this observer also this light pulse will be travelling with speed c this light pulse will also be travelling with speed c. Remember in the classical mechanics in the classical velocity transformation the speeds were different, but if the second postulate of special theory is correct then this speed must be c this speed also must be c because the speed of light is a frame independent quantity both the observers will feel exactly the speed to be identical. So, according to the ground observer this pulse which was which originated from this particular center was actually travelling with the speed c the one which was going backwards also travelled with the same speed c. But by the time the pulse light pulse of light moves from this particular position towards this particular wall this wall has moved ahead because the train is actually moving. So, it takes certain amount of time for the light to reach from this particular point to this particular point and according to the ground observer during this particular time this train has moved ahead and gone somewhere here. So, by the time this light will go and hit this particular wall it has to travel a larger distance. On the other hand this pulse when it was trying to move towards the back wall this back wall was approaching towards this particular source of light. So, eventually this light has to travel a smaller distance before hitting this particular wall. So, remember the same concept was used also by galleon observer when we are throwing the balls the only difference in that particular case was that according to that observer this ball was thrown with a larger speed and this ball was thrown with a smaller speed. But now both the light pulses travelled with the same speed if the second postulate is correct. So, what then observer on the ground will conclude the observer on the ground will conclude that this particular light will reach this particular wall later than here because both the light pulses are travelling with the same speed, but this travels a smaller distance to reach the wall while this travels a larger distance to reach the wall. Obviously, these two events cannot be simultaneous. This event number 1 will occur later than the event number 2 that is reaching this particular wall. This is what I have written here. The two events again would have to turned out to be simultaneous if we had used the classical velocity as in the case of balls. However, under the second postulate the speed of light is still see in both the directions, but it has to travel a larger distance to reach the front wall then the back wall as just now I have explained. Hence, event 2 occurs before event 1. So, the two events are not simultaneous. So, simultaneity is also related. Now, let us come back to our discussion on time. See, time is often related to the simultaneity of two events. See, when I say that a particular event happens, let us say class is starting and taking another event or let us say class is starting at let us say 9 o'clock. Essentially, we mean that when watch shows 9 o'clock and when the class starts, these two events are simultaneous. Or when we say that a train starts from a given station at let us say 10 o'clock, it means when watch shows 10 o'clock and when the train starts, these two events are simultaneous. We have just now seen that simultaneity is relative. It means there could be another frame in which these two events may not appear to be occurring simultaneous or appearing to be occur at the same time. It means we probably will have to question the same basic equation which we have written if we have sort of implicitly assumed so far which is t is equal to t. So, this is what I have written here. Time is often related to simultaneity of two events. We have just seen that simultaneity also depends on frame under the second postulate. Therefore, probably we have a reason to believe that t prime is not equal to t or t is not equal to t prime. So, probably one has to relook at the concept of time and we cannot use a simple equation like t prime equal to t. So, now we are again back into dark. We do not know where to start. All we have said that probably time cannot be taken to be same in all the frames. So, let us start collecting all our evidences and try to reach a new transformation. This transformation is called Lorentz transformation. So, let us just before starting let us make some observations. We are as I said we are essentially in dark. So, we have to start collecting all the clues that we can get. I must mention that we are not deriving Lorentz transformation. These things are not derivable. So, when we say something to be derived it means we can derive it from certain basic fundamental laws. So, we are only collecting arguments. We are not deriving it. These observations whatever we have seen whatever we feel should be put into the in a set of equations. But whether these equations are correct or not depends on whether they can explain the experiment. Like when we say Newton's law of motion. Newton's law of motion we do not derive in our high schools. We are never told how to derive Newton's law of motion. They became a fundamental law. We assume Newton's law of motion and try to explain various things and that is the way we had developed our physics. Similarly in Lorentz transformation we are just collecting our evidences. Some of these evidences may not look very strong. Some of these observations may not look very strong. But eventually we believe in them because they are able to explain a large number of experiments which we cannot otherwise explain. That is why we believe in Lorentz transformation. So, the first point about Lorentz transformation is that the transformation should be linear. I must mention that the thing that I am taking this particular treatment is given in Resnick's book which I will be the reference will be I will be giving in the next transparency. So, what I mean by a linear transformation if you see this particular equation we have written x prime is equal to one constant which I am calling as Bxx because this is relating x to x. This is Bxy relating x to y Bxz Bxt and C1. So, we have written x prime is equal to Bxx x plus Bxy y plus Bxz z plus Bxt t where all these B's are some constants which we have to determine. By linear mean we mean that there is only one single power of x which is involved there is single power of y which is involved single power of z which is involved single power of t is involved. We do not have a term involving let us say x square or x cube or y square or term like x y or term like z t. We expect the transformation to be linear. I will give you the reason why do we expect it to be like that. You look at y prime we have written exactly similar type of equation except these constants are different which in general would be different. This I have written as Byx this y represents this y this x represents this x Byx x plus Byy y plus Byz z plus Byt t plus C2. Similarly, z prime we have written as Bzx x plus Bzy y plus Bzz z plus Bzt plus constant C3. Then time we have written which earlier was always equal to t t prime is equal to Btx x plus Bty y plus Btz z plus Bt t t plus C4. Let me first give an argument why we expect this particular transformation to be linear. Let us suppose we had a transformation which involved x square term. If we had a transformation which involved x square term and let us suppose let us just put a rod which is of let us say length 1 meter which is placed between x is equal to 0 and x is equal to 1. So, if this rod is put between x is equal to 0 and x is equal to 1, then x 1 will be equal to 1 and x 0 will be equal to 0. Now, x 1 minus 0 will be equal to 1. If I take x 1 square, this will be 1, x 0 square will be equal to 0. Now, let us suppose we displace this rod and put this particular rod now between x is equal to 1 and x is equal to 2 same 1 meter rod. Now, let us call this x 3, x 3 will be equal to 1, x 3, let me put it x 4 will be equal to 2, x 3 will be equal to 1 because this is now at x is equal to 2 meters and this is at x is equal to 1 meter. If we take x 4 square, this will be 4. If we take x 3 square, this is equal to 1. So, if my transformation term involved x square, we will find that x square, x 1 square minus x 0 square is 1 while x 4 square minus x 3 square is 3. So, this particular length would appear to be 1 meter in S frame of reference if we had a term like this and this particular length will turn out to be 3 meters. It means if this rod was put between x is equal to 0 and 1, its length will turn out to be different from the case when the same rod was put between x is equal to 1 and x is equal to 2. This looks somewhat bizarre. This does not look somewhat reasonable that if we just place our rod and displace it by 1 meter, the length of the rod in a different frame will appear out to be different. After all whatever we have taken as origin could have been a slightly different origin. We expect the space to be homogeneous. I do not expect the length of a rod to be different in different frames based on where I put the rod. If the length I can accept that you know the length of 1 meter may not appear to be 1 meter, it may appear to be 1.2 or 0.8 meter. But if I displace the rod, the same length 1.2 or 0.8 whatever we had observed earlier, same length should be observed in S frame of reference. By displacing the rod, I should not be able to, I should not measure a different length because the space is homogeneous. I do not expect length to be dependent on the fact where I put this particular rod. Suppose this is particular pen, if I displace it here, it should not appear that in a different frame of reference, its length has changed. Whatever might be the length larger or smaller, but whatever was the length here, same should be the length observed here also. So, therefore, I avoid all higher order terms and we expect linearity. So, the first clue that we have that transformation must be linear, it must involve only single powers, it should not involve powers like x square, x cube, etc., etc. So, the simple most transformation in linear is of this particular type which we have just now mentioned. So, this is what I have said linearity is essential to maintain homogeneity of space, the length of the rod should not depend on the origin chosen. And this is the reference of the book from which I have taken this particular derivation. This is Introduction to Special Relativity by Robert Resnick by Boille Eastern, 1988. Now, let us collect more evidences. What we will realize, see earlier when we introduced Galilean transformation, we had arbitrarily put our S and S prime frames and taken essentially a simplistic view of the axes. Now, you will realize that this simplistic view helps us in getting eliminated a large number of constants and therefore, making transformation equals simple. I remind again your two frames S and S prime, this was the origin O, this was the origin of O prime, this was my x axis, this was y axis, this is z axis, this is z prime axis, this is y prime axis, x and x prime axis are also always coincident and the time t is equal to 0 was measured when O prime was coincident with O. These are the special axes that we had taken for our Galilean transformation. Now, let us see that by checking these set of axes, we are ready to get, we can eliminate a large number of constants. So, this is what I have written, special choice of axes make many coefficient 0. First, making this particular origin, choosing these origins appropriately. Remember, we have chosen these origins that O prime when it coincides with O, then at that particular moment, the time is measured to be 0. So, let us assume that at a time when the origins of the two frames were coincident and event occurred, whatever might be the event, it occurred exactly at the origin. So, it means at that particular when the event occurred, the coordinate of that particular event in S frame was x is equal to 0, y is equal to 0, z is equal to 0, because it occurred at origin and at time t is equal to 0. Now, because of the special choices that we have taken, we expect that this event in S frame of reference would also occur it is at origin, because it was coincident at that particular time and because time was also measured from that particular instant of time. So, time in that particular frame will also turn out to be 0. So, I expect that in S frame of reference, also the event will turn out to be at 0, 0, 0 and at time t prime equal to 0. So, let us put these conditions in this particular transformation equation which we have just now written and let us see that we can get rid of some constants. So, I have written let us imagine that an event occurs at origin in S at t is equal to 0, the event would also appear to occur at the origin of S prime at t prime is equal to 0. It means if I substitute x is equal to 0, y is equal to 0, z is equal to 0, t is equal to 0, then I must get x prime equal to 0. So, remember if you look at this equation, this will give you now 0 is equal to 0 plus 0 plus 0 plus 0 plus c 1, which means that this c 1 must be 0, there cannot be a cos center. Similarly, we can put these conditions in the y z coordinate system also, y z transformation equation also and we will conclude that c 2 is also equal to 0 and c 3 is also equal to 0. Similarly, for time if the event occurred at x is equal to 0, I put it 0 here, occurs at y is equal to 0, I put a 0 here, occurs at z is equal to 0, I put a 0 here, it occurs at t is equal to 0, I put it here, then the watch of S prime observer must also measure time t prime equal to 0. So, if I put this equal to 0, I will get c 4 also to be 0. So, I conclude that by choosing these special set of axes, these special conditions on the two frames, I am able to get rid of the constant c 1, c 2, c 3, c 4. So, this is what I have written here and the transformation equation now reduced to these. So, remember we had so many constants out of this c 1 has gone down, has become 0, c 2 has been removed, it has become 0, c 3 has been removed, it has become 0, c 4 has been removed, it has become 0. So, now we are still left with 16 constants, still a very large number of the constants and let us see if we can get rid of some other constants. Now, let us start looking again at the axes and trying to see how these special choice that we have made about the axes can help us get rid of some more constants. So, we try to fix the planes appropriately which we have already done by making these special choices which is given here. Let us now imagine that an event occurs at arbitrary time in x y plane. So, remember this is your x y plane, this is your x direction, this is your y direction. So, this is x y plane. Now, if we realize that in s frame of reference, this x prime axis is coincident with x, y prime axis is parallel to y. So, we expect that this x prime y prime plane is exactly same as x y plane. They have the same planes because in principle, this particular line moves in that in that particular plane, the y prime direction moves in the plane. So, x y plane is same as x prime y prime plane. It means that if an event occurs in x y plane, an observer in s frame of reference will also observe it to be in his x prime y prime plane. Now, what is x y plane characterized of? x y plane is characterized by the fact that z is equal to 0. x y plane means z coordinate is 0. Now, x prime y prime plane also means z prime equal to 0. So, what we conclude from this particular discussion? That if an event has occurred with z equal to 0 means in x y plane, it would appear to an observer in s frame also to occur at x prime y in the x prime y prime plane, it means according to that observer, z prime will also be equal to 0. So, let us put this condition in our equations and let us see what happens. So, this is the third equation in which I have put z equal to 0 here and I expect that if z is equal to 0, I must also have z prime equal to 0. Now, this is possible only and remember this event could occur at any arbitrary value of x y and t, but so long it occurs in with a coordinate z equal to 0, the coordinate of that same event must also have z prime equal to 0. So, irrespective of value of x y and t, this equation must always be true. That is only possible if this coefficient is 0, this coefficient is 0, this coefficient is 0, this coefficient is 0. So, in one shot we get rid of these three constants. So, this is what I have written this is possible only if v z x, v z y and v z t all are set to 0, then only it is possible that an event which occurs with z equal to 0 will also appear to an observer in s prime to occur with z prime equal to 0. Similarly, now we can look at x z plane. So, let us come back to this particular equation. Remember x z plane, this is the x z plane. We also realize that x z plane is same as x prime z plane, z prime plane. So, if an event occurs in x z plane implying y equal to 0, then the event would also appear in s prime frame of reference to occur in x prime z prime plane. It means it must occur with y prime equal to 0, exactly the same condition that we had applied for x y plane. So, we put exactly the same thing, same condition that corresponding to y equal to 0, you must have y prime equal to 0. Therefore, you will get rid of this constant, this constant, this constant. It is exactly the same argument that we had used for x y plane. So, we get rid of another three constants. Now, you remember there is one plane which is different and that is y z plane, this plane y z plane and y prime z prime plane. These planes were coincident only at t is equal to 0. At a later time, these planes though they are parallel, but they are moving relative to each other. So, it was different for x z and x y planes, which always remain identical in the two frames. But as far as y z plane is concerned, they were identical exactly at t is equal to 0. But at a later time, they have moved from each other and how much they have moved? They have moved by a distance of v times t, where v is the relative speed between the frame. So, an observer in s frame would find that this particular y prime z prime plane has moved with respect to y z plane in a time t with a speed v. So, the distance between them is v t. So, let me repeat, an observer in s frame would find that the y prime z prime plane has moved with respect to y z plane by a distance of v t during time t. This time t is being measured in s frame. All the observations are being made in s frame. Let us see what does this put? This puts a condition on our constants. So, this is what I have written. Let us look at the y z plane. At the time t is equal to 0, if an event occurred in this plane, it would also appear to occur in y prime z prime plane to observe in s prime. But at a later time, the x coordinate of this particular event will be shifted by v t as observed in s frame. Therefore, the transformation equation must look as follows. We have taken s quite a bit of jump here, which is x prime is equal to b x x into x minus v t. Because if I put x is equal to v t, then I must get x prime is equal to 0. Remember this equation. If I put x prime equal to, if you put x is equal to v t, then I must get this point in this particular y prime z prime plane. And other constants must turn out to be 0. So, we find this constant to be 0. We find this particular constant to be 0. And x prime we expect to be of the form v x x multiplied by x minus v t. So, as we can see that these special choice of the axes have led us into a large number of a large amount of simplification getting rid of so many constants. So, let us clean up all our equations. And we obtain the following equations. x prime is equal to b x x into x minus v t, y prime is equal to b y y times y, z prime is equal to b z z times z, t prime is equal to b t x multiplied by x plus b t y multiplied by y plus b t z multiplied by z plus b t t multiplied by t. We still have 4, 6, 7 constants. Let us see how we are going to evaluate these constants. But I hope you are able to appreciate that the special set of axes that we have chosen, though they are not really in that sense special, they are general enough. But just by choosing those set of axes, we are able to simplify a large amount of, we are able to simplify our transformation equations. Otherwise, we do not know how to go about it. Now, let us look at the symmetry arguments. The symmetry arguments are very interesting arguments, which often can work out very well in physics. See, when I am talking about the x axis, x axis is sort of a well-defined axis because x axis is the direction of the relative speed. See, suppose you are just looking at the motion of two frames and you have to define the relative velocity direction. You have to define the x direction. How we will proceed? We will proceed by taking the direction of the relative motion is x direction. But once we have chosen x direction, the y direction, whether I took this way or whether I took this way or whether I took at another angle, it does not make a difference. For example, if I have taken this as my x axis, I could have taken this as my y axis or I could have taken this as my y axis, making an angle, so long it is perpendicular, I can take this as y axis. y axis was under my control. There is no reason of specifying or choosing a particular specific y value, y axis. x axis is definitely, x axis is the direction along which the relative velocity occurs. This is the v direction, which is a unique direction along which the sort of symmetry is broken because that direction is a unique direction defined by the relative velocity directions. But the same thing is not true about y and z. I can choose y axis any way I like. Of course, it has to be perpendicular to x, then between x and y, z axis has to be perpendicular to x axis and y axis and must follow the right handed rule, whatever is the standard rule for choosing the axis. But on the other hand, this y axis could have been anywhere along this particular plane. So, I do not expect my physics laws to change depending on what I choose as my y axis. So, there has to be symmetry in that particular direction. So, let us imagine now, let us imagine that we have chosen some particular y axis, correspondingly we have chosen z axis. So, my minus y prime axis minus z prime axis, all these axes are known. Now, let us suppose an event occurs at x naught, y naught, z naught and x naught minus y naught, z naught. This is your x direction, this is y direction, this is z direction. As I say, x direction is a unique direction, which I cannot change because that depends on the relative velocity. y direction I could have chosen anything, I have chosen one particular thing, which is happening, which is along this particular direction. Let us suppose anything, the two events, which are occurring here and exactly with a negative value of y. So, this occurs at x naught, y naught, z naught, this occurs at x naught minus y naught and z naught. So, all that is difference between the coordinates of these two events is the y coordinate. Otherwise, all the three other two coordinates are same. Now, let us look back at our time equation, which we have written earlier, which is t prime is equal to b t x x plus b t y y plus b t z z. Now, if I change y to minus y and if b t y is not 0, then t prime will turn out to be different for different frames. This y changing sign of y will change the time. But after all, there is nothing special about this particular y. What I have called this y as positive, I could have chosen my y axis just opposite to that. All right, of course, my z axis will also change, but this y axis could have been just taken as negative of that and that event would have now appear to occur at minus y naught. What event appear to be occurring at plus y naught may have occurred at minus y naught, because this just depends on us how I choose the axis. So, I do not expect that physics will change, because just I have taken a different set of y axis. So, the time of the event, the other person does not know what is the choice of the axis. His time should not turn out to be different depending upon what I have chosen as my y axis or what I have chosen as my minus y axis. So, I do not expect that by reversing the coordinate y naught, I would expect that time to be changing. Similarly, if I assume two events to be occurring at exactly the same value of x naught, y naught and only z naught being different and occurring at x naught, y naught minus z naught, I do not expect that time to be different, because these are essentially symmetrical arguments. The coordinate that I have chosen have just depend upon one particular choice for which I did not have any prior information how to choose that particular direction, that was totally arbitrary. Therefore, just by taking the choice of the axis, the time of the event should not change as being observed in S frame of reference. So, remember I remind along the x direction is a unique direction, x direction things can change, because there is a special reason to choose the x direction, but not for y and not for z. So, I expect that these two coordinates, these two constants b, t, y and b, t, z must be equal to 0 in order to maintain the symmetry. So, this is what I have written. Imagine two events occurring at the same time in S frame, let the coordinates of the two events be x naught, y naught, z naught and x naught minus y naught, z naught. If t prime depends on y, then time of these two events would appear to be different in different frame, but what we call y axis could have also been termed as minus y. The choice of x axis is unique as it is determined by the direction of the relative velocity in our choice of axis, but not of y and z. Hence, the transformation equation must appear as follows. It means I must have these two terms also becoming 0. Therefore, I get rid of two further constants. So, now we are left with only 1, 2, 3, 4, 5 constants. This is what I am left with. As we can see that we have simplified reasonably well purely by physics arguments and special set of axes transformation equations to a simple form. I would just like to remind you that Galilean transformation is also a special case of these equations. Remember in Galilean transformation, v x s was 1, d y y was 1, b z z was 1, b t x was 0, b t t was 1. So, Galilean transformation also is a special case of this. Remember till now we have not involved any special, any postulates of special theory of relativity. This transformation equation whatever we have discussed are general enough so long we take the same set of axes that we have chosen. So, this is expected to be generally true whether it is a classical rule or under special theory of relativity. We expect the same equations to be true. So, let us go to the summary of whatever we have discussed today. We discuss using examples that we have to attack the equation which makes the time same in the two frames. We then discuss the form of transformation equations without invoking any postulates of relativity only making time relative. Only we have taken consider the fact that this time could also be relative. We have not invoked any things of special theory of relativity. So, in our next lecture we will invoke now the conditions of special theory of relativity and will determine all the constants that we have not so far determined and will arrive at what we call today Lorentz transformation.