 So, let us recapitulate what we did last time. We discussed the simultaneity of two events that was the first thing that we discussed. We tried to see that if there are two events which occur at the same time in a given frame of reference, which we normally call as simultaneous events may not appear to be simultaneous in a different frame or in general they would not appear to be simultaneous in another frame. So, this is what we say is that simultaneity is relative. This will be true if second postulate is correct. Then we also started collecting our arguments to look for a new transformation because we realize that Galilean transformation would not be consistent with the second postulate. So, we started collecting all our arguments in the lookout for a new transformation in which time has also to be made frame dependent. In fact, we gave reasonings that why we expect time also to be frame dependent. After that, we started looking for a new transformation without invoking any of the postulates of special theory of relativity. Just purely on physics grounds, we try to look for a transformation which we try to look at the form of the transformation the way it should look like. And we also realize that Galilean transformation is also a special case of that. So, without using any of the postulates of special theory of relativity, we arrived at the form of these equations to which Galilean transformation is a special case. The transformation equations eventually turn out to be of the following form which is written here. See, x prime is equal to bxx multiplied by x minus vt, where bxx is a constant which is yet to be evaluated. y prime is equal to byy multiplied by y, again byy is a constant to be evaluated, z prime is equal to bzz times z, t prime is btxx plus btt. We realize that this transformation similar to the Galilean transformation in which bxx was 1, byy was 1, bzz was 1, btx was 0 and btt was 1. So, by taking special values of these constants, Galilean transformation can be arrived at. Now, we have to evaluate these constants by invoking the postulates of special theory of relativity. That is what we are going to do just now. See, first postulate that we try to invoke here in order to evaluate these constants is the fact that we expect all inertial frames of references to be equivalent. We do not expect any preferential inertial frame of reference. Now, we realize that if you go back to our equations and we take y prime is equal to byy multiplied by y as this equation suggests, then let us suppose we put a rod in S frame between 0 and 1. So, we have a rod which is we are putting and here it is y is equal to 0, here it is y is equal to 1 as measured in S frame. I am drawing the same rod here, y is equal to 0, y is equal to 1. This is my x axis, this is my y axis and let us assume that this is put along the y axis. Now, according to this particular transformation law which is y prime is equal to byy times y, y is equal to 0 would give me y prime is equal to 0, y is equal to 1 will give me y prime is equal to byy. So, from this an observer in S prime would conclude because this is the transformation equations which are supposed to translate information from one frame to another frame. So, according to this transformation equations if they are correct, the length of that particular rod would be found out to be byy by an observer in S prime because according to him the rod is put between y prime is equal to 0 and y prime is equal to byy. Now, let us imagine a different situation a inverted situation in which the rod was actually put in y S frame of reference between y prime is equal to 0 and y prime is equal to 1. Exactly the same thing except that now the rod is put in S prime frame of reference and put between y prime is equal to 0 and y prime is equal to 1. So, this is my rod. Now, if these transformation equations are correct the one which we have mentioned we apply the same transformation here. So, y prime is equal to 0 would give me y is equal to 0 no problem, y prime is equal to 1 will give me y is equal to 1 divided by byy. Now, if byy happens to be anything different from 1 then in that particular case this observer here would say let us assume that byy is let us assume that byy is equal to 2. So, this observer here would assume would find out that the length of the rod is 2 meters, but if the same rod was put in y prime frame of reference the observer in S frame will find out that the length of the rod is only half meter 1 divided by 2. It means one frame of reference is magnifying the length of the rod another frame of reference is reducing the length of the rod. This we do not expect. See, I do not mind that in y frame if I put a rod along y direction or any direction its length turns out to be reduced in y prime frame of reference or S prime frame of reference, but the inversely should also be true. It means if exactly in a similar location in S prime frame of reference we put a rod in S frame also it should turn out to be reduced that is what is meant by the equivalence of the two frames. It cannot happen that in one frame all the rods appear to be elongated in comparison to another and when we inverse the frame and the other frame they will turn out to be all reduced. It means they are not equivalent. I can distinguish by knowing the length of the rod whether this frame is different from the other which is not allowed by the first postulates of special theory first postulate of special theory of relativity. Hence, I expect that byy must be equal to 1 because unless it is equal to 1 I will be able to distinguish I will evolve a method of distinguishing between two frames one as magnifying frame is another is reducing frame. Hence, I expect byy to be equal to 1. So, this is what I have written in the next transparency. We first use the fact that there should be no preferential frame if you put 1 meter rod between y is equal to 0 and y is equal to 1 meter its length in S prime would appear to be byy as per transformation. The same rod is put between y prime is equal to 0 and y prime is equal to 1 meter its length would be 1 divided by byy meter just now the way I have discussed. This is not expected as the frames become distinguishable as magnifying or reducing. Therefore, we expect byy to be equal to 1. Now, exactly in a similar way I can put the rods along the z direction and using exactly the same argument I can put bzz equal to 1. So, therefore, I must have byy is equal to 1 and bzz is equal to 1. So, two unknown constants have now disappeared we have we are now left with three constants. Remember same argument I cannot apply along the x direction because in x direction there is a time dependence as well as x dependence. So, things are little more difficult as far as the x direction is concerned. Remember x direction is somewhat unique because this is determined by the direction of the relative velocity while as I have discussed in last lecture why direction could have been rotated with respect to x direction and things would not have changed. So, now my equations become comparatively simpler with only three unknowns which is x prime is equal to bxx multiplied by x minus vt y prime now becomes equal to y y z prime is equal to z and t prime is equal to btx x plus bttt. So, these are the four equations in which there are three unknowns bxx btx and btt which have to be still determined using the postulates of special theory of relativity. Now, this is the time I will involve the second postulate of special theory of relativity to determine the three constants. Let us assume that at t is equal to 0 when the origins of the two frames are coincident a particular source of light emits a light from origin when we can always imagine like that. Had it been classical mechanics we would have always asked the question suppose we are asking a question a ball is being thrown from the origins we will always ask the question whether this person who was throwing the ball was he stationary in S frame or was he stationary in S prime frame or was he stationary in any other frame but for light we need not answer this question because whatever might be the frame according to the second postulate the light speed will always be c. So, it makes no difference whether the source which was emitting light was stationary in S or S prime or in any other frame all I am saying that the incident when the two frames were coincident at that instant which is time t is equal to 0 and t prime is equal to 0 there was a source of light at the origin which emits light. Now, this light let us assume an observer in S frame according to that particular person this light will emerge in a spherical wave front which will appear as a sphere with the center with the origin as the center of the sphere. In all the direction it has to travel with the speed of light c as observed in S frame of reference. In S prime frame of reference exactly the same thing would happen that person would also feel that this light was actually originated from the origin of his frame and in all the directions this spherical wave front is emerging with his center as origin with his origin as a center that is what the observer in S prime frame will also feel. So, he will also feel remember at a given later time the origins of the two frames of reference are no longer coincident, but still both of them will be feeling as if the spherical wave front is emerging with their origins with their respective origins at the center of the sphere. This is what I have shown in this particular transparency which is something like this. This is sphere this another sphere so it appears to be emerging according to the observer in this red frame of reference in S frame of reference this wave front appears to be spherical with its center at the origin. According to this observer also this particular wave front would appear to be spherical with this O prime as origin. Remember at a given time O and O prime are not coincident still both of them will feel as if it has been emerging from their own centers. This is what I have written here at T is equal to T prime is equal to 0 a spherical light wave is emitted from the origin which is going in all the directions that is what we mean by spherical light wave. The observer in both S and S prime will find that the spherical wave front is emerging from their respective centers with the same speed C in all the directions. Therefore, if I have to write the equation then observer in the equation of the spherical wave front according to the observer in S frame this equation will be an equation of a sphere the radius of which is changing as a function of time and at a given time the radius is given by C multiplied by T where C is the speed of light. So, it will appear to be x square plus y square plus z square is equal to C square T square that is the equation of the spherical wave front which observer in S will write. On the other hand observer in S prime remember he has to be consistent in his frame will write an equation which is x prime square plus y prime square plus z prime square is equal to C square T prime square because according to that particular observer the radius of the sphere is given by C multiplied by T prime where T prime is the time in his frame x prime y prime z prime are the coordinates measured in his frame. So according to S this equation must be correct as the equation of the spherical wave front according to S prime equation observer this should be the equation of spherical wave front. So whatever transformation equations that we have written if I substitute those transformation equations here in this particular thing because transformation equations are supposed to translate information from one frame to another frame. So if these transformation equations are correct if I substitute this transformation for x prime y prime z prime in these equations and of course T prime then I must get back these equations this second equation that is correct then I have found out a correct transformation equation. So remember we have already looked at our transformation equations let us substitute it back in this particular thing and try to see whether we can evaluate the other three remaining constants which we have not done so far. So this is what I have written substitute information transformation equations obtained so far in the second equation and if we substitute we will get the following. This is the equation which observer in S prime wrote x prime square plus y prime square plus z prime square is equal to c square T prime square. If you remember we have reached up till this particular point as far as transformation equations are correct x prime is equal to bxx multiplied by x minus vt y prime is equal to y z prime equal to z T prime is equal to btxx plus bttt so I substitute back in this equation. So for x prime I write this thing this is what I have written so once I substitute this thing here in this equation this whole thing will get squared so this becomes bxx prime square square of this x minus vt so square of x minus vt so I have just put x prime square about y prime there is no problem about z prime there is no problem because y prime happens to be equal to y z prime happens to be equal to z so if I substitute it back here I just get y square z square then on the right hand side I have c square which anyway does not change with frames we have T prime so what T prime I put this equation which is btx plus bttt so on the right hand side I will get c square which I have put here c square T prime square it means the square of this particular quantity so I have put bttx plus bttt whole square so all I have done is these trial transformation equations have been put in x prime y prime z prime equation this equation and I want to see that I arrive back at my first equation which is x square plus y square plus z square is equal to c square T square if that happens to be true then only my transformation equations are correct. So let us go to the next transparency and try to expand this this is the equation which I have just now written in my earlier transparency I am expanding this squares so this is bxx square which remains x minus vt I expand this equation you know the standard equation for a plus b whole square is equal to a square plus 2ab plus b square I use exactly the same thing here so my a is x my b is vt so a square which is x square minus 2ab so there is a 2 multiplied by x multiplied by vt plus b square so v square t square plus y square plus z square is equal to c square which is here then I expand this square so this becomes btxx whole square that is a square so this is btx square x square 2ab so 2 this multiplied by this so 2 btx I have taken this btt first so this becomes 2 btx multiplied by btt into x multiplied by tf slightly reorganize this term then b square which is the square of this particular term so this becomes btt square plus t square so all I have done is expanded these square quantities to write this equation and I want to now pick up x square y square z square and try to see that it matches with the original equation x square plus y square plus z square is equal to c square t square so let us do that in the next transparency this is the equation which I had written in the last transparency that is the beginning equation now I am starting I have started picking up terms involving x square so this is x square so this term contains x square so this is btxx square so this is btxx square let us see which term other contains x square it does not contain x square this does not contain x square only other term which contains x square is this which is on the right hand side of equal sign so I take it on the left hand side so this equation becomes negative so becomes minus c square multiplied by btx square multiplied by x square so this is what I have written here so this is the second term which contains x square so the coefficient of x square becomes btx square minus c square btx square no other term contains x square y square straight forward y square only z square there is no problem it is z square then I am trying to collect 2xt term there is x and t here so I am trying to collect 2xt term which is here I am taking negative sign out here so this term becomes positive the coefficient of 2xt will be v multiplied by btxx square so this I have written here btx square multiplied by v this is this term there is another term which contains 2xt which is on the right hand side of the equal sign so once I bring it back it will become negative a negative has already been taken out so this term becomes positive so it becomes remember we are looking at the coefficient of 2xt so it becomes btx ptt multiplied by c square this is what is here no other term contains x and t together now let us look at the t square term t square term there is one t square term here the coefficient of t square term is bx square plus v square but I want to take this on the right hand side of equal sign because it is equal we are putting this on the right hand side so this term would now become negative so bxx square multiplied by v square will become negative so that is what I have written here bxx square v square as negative there is already one term of t square on the right hand side which is btt square multiplied by c square so this remains positive btt square c square minus vxx square v square so this is simply collection of all the terms second order terms involving x square y square z square xt and t square I have put the equation back here exactly the same equation which I have put in the last transparency remember if my transformation equations are correct then this equation should exactly look like x square plus y square plus z square is equal to c square t square that is what we have been discussing just now if these equations have to match it means this coefficient must be equal to 1 y square z square anyway there is one no issue 2xt there is no term of 2xt here so I cannot allow a 2xt term here it means this coefficient must be 0 then only there will be no 2xt term this t square coefficient in this particular equation is c square therefore this coefficient this particular term must be equal to c square so I get 3 equations out of this one is this whole term should be equal to 1 this whole term should be equal to 0 this whole term should be equal to c square so if whatever we have said is correct then these 3 equations must be valid and remember we had only 3 unknown constants and if we have 3 equations we can determine these coefficients and my transformation equation becomes known this is what I have written in the next transparency so you have bxx square minus c square btx square is equal to 1 btx bttc square plus bxx square v is 0 btt square c square minus bxx square v square should be equal to c square with these 3 equations 3 unknowns can be determined only one thing one has to take care because there are so many quadratics involved here once you try to solve you will find out that there may be 2 roots possible for some of these terms okay so you have to properly take care of sine how to take care of sine I will just explain in a minute let us go back to the equations which we have just now written in fact these are the solve sort of solutions but before I discuss that let me just discuss this particular point here I expect this coefficient to be positive even though there may be a negative root which may be allowed as a solution this bxx has to be positive the reason is that that at time t is equal to 0 f in event occurs at a positive value of x in s frame I expect it also to occur at a positive value of x prime in s frame of reference if bxx was negative the x's will change sine at t is equal to 0 which we do not expect because it should not happen in one frame and I do not physically expect that something which is happening at positive values of x okay and will happen at negative values of x in a different frame it is sort of reflection we do not expect therefore I expect bxx must be equal to positive exactly in a similar condition we must also expect that the coefficient of btt must also be positive because at x is equal to 0 whatever is the time sequence if it is positive if it is a positive time it should also appear to be positive in s frame of reference and if you look back at this particular equation if this has to be positive this has to be positive this btx must be taken negative so that this equation must be equal to 0 so these are some of the sine kr1 has to take in order to arrive finally at the solutions and these are the solutions that we get bxx is equal to btt is equal to 1 divided by under root 1 minus v square by c square for btx I will get negative as I had said that should be negative minus v divided by c square divided by under root 1 minus v square by c square so I have found out all the three coefficients and hence I have found out the transformation which normally we call as Lorentz transformation this is called Lorentz transformation this particular transformation it appears was derived by Lorentz with slightly different meaning in a different context that is why this particular transformation is named as Lorentz transformation so these are my Lorentz transformation equations where I have put all these constants back so x prime is equal to gamma minus vt divided by under root 1 minus v square by c square these two are simple equations t prime is equal to t minus vx upon c square divided by under root 1 minus v square by c square normally these equations tend to become somewhat clumsy so we use some notations which is generally very standardly used special theory of relativity v upon c we call as beta so beta is equal to v divided by c and gamma we define as 1 divided by under root 1 minus v square by c square we have already defined beta as v by c so this becomes 1 minus beta square so I normally use gamma as 1 divided by under root 1 minus beta square looking at this particular look using this particular abbreviations the equations appear somewhat simple to write x prime is equal to gamma times x minus vt y prime is equal to y z prime equal to z and t prime is equal to gamma multiplied by t minus vx upon c square so these are my Lorentz transformation equations. Before we start discussing these equations let me make some observations on Lorentz transformation which are fairly interesting and probably try to bring out some similarity or some dissimilarity with the traditional or classical Galilean transformation. First thing that we observe that if the relative speed v between the frame is comparatively very small this particular transformation would reduce to Galilean transformation it is easy to see because v is very small in comparison to c it means beta is negligible if beta is negligible in comparison to 1 then gamma is essentially very close to 1 and if gamma is very close to 1 then this becomes x prime is equal to x minus vt which is the x coordinate transformation equation of Galilean transformation. If I look at this particular term gamma is 1 v being very small in comparison to c this term can also be neglected and this equation also becomes t prime is equal to t which is the traditional Galilean transformation equation for time which assume that time is same in all the frames t prime equal to t. So we realize that in the classical limit which is v much smaller in comparison to c the Lorentz transformation is reducing to Galilean transformation. This is something which we expected because in normal life for example if a car is moving or a train is moving or a airplane is moving we never see relativistic effects the effects that we are going to describe later. So we definitely expect that the speeds when the speed that we are talking if they are not really that high as comparison to speed of light we mean we should not expect relativistic effects to be observed and that is what we normally do. So we do find that in the classical limit what we call as a classical limit that v much smaller than c you will always reduce back to Galilean transformation. So if we are not talking of very large speeds Galilean transformation is alright which is simple. That thing is probably an interesting comparison between Galilean transformation and the Lorentz transformation. See let us imagine a situation when an event occurs at origin in S. Let us suppose this is an observer S and some event occurs at the origin. So an observer is sitting at the origin some event let us assume that a train just passing by the side or anything. So a particular event occurs just at the origin. Now my question is that will to an observer an S frame of reference would it also appear to occur at origin? Classically you can answer this particular question it need not be because it depends where the origin of that particular observer in S or origin of S frame where is it at that particular time. So if the origin was here S frame of reference origin was here then to this particular observer this event would be appearing to occur at a different value of X. Unless the time for this event was also 0 the time of this particular event was 0 in that particular case this origin was coincident at this origin remember our special conditions on the axis and at that time the observer in S frame of reference would also find that this particular event occurred at the origin. Let me just write the Lorentz transformation equation again we have to keep on referring it quite often t prime is equal to gamma t minus v X upon c square remember gamma let us not be confused with this this is gamma this is y this is y prime is equal to gamma this gamma is also dependent on the speed of light these are my Lorentz transformation. Now same thing is seen in the Lorentz transformation also if the event occurs at X is equal to 0 the same event would appear to occur at X prime equal to 0 only if t is equal to 0 if t is not equal to 0 this X prime may occur may have a different value from 0 this is something which is classical and it happens generally to we all know there is nothing surprising about it Lorentz transformation also gives the same thing why I am telling something so obvious is only because I want to compare with the second statement which I am coming just now. So, let me just read whatever I have written here if an event occurs at X is equal to 0 in S it would appear to occur at X prime is equal to 0 in S only when the event occurred at t is equal to 0 at a later time the origins are no longer coincident therefore it may not appear to be occurring at X prime is equal to 0 this is true even classically. But let us look at this time equation if an event occurs at t is equal to 0 in S frame same event you know for example train passes or let us assume train is passing with a very high speed and relativistic speed very close to the speed of light it is passing by the origin at t is equal to 0 it appears to be passing now at t prime is equal to the observer in S prime would also find it out that this is at same time t prime is equal to 0 if this event occurs only at X is equal to 0 if for example this train was passing at a distance of 20 kilometers away from the origin in S frame at time t is equal to 0 then S prime observer would not find this particular train to be this particular event of train passing happening at t prime is equal to 0 or let us assume that there is a lightning which happens at 20 kilometers away from the origin at time t is equal to 0 the observer in S frame of reference would not find that this particular lightning occurred at the same time look at this particular equation t prime is equal to gamma t minus v x upon c square if t equal to 0 that does not garden t t prime equal to 0 unless X is also equal to 0. If X is not equal to 0 t prime is different from t so if an event occurs at S at 0 time then that event may not appear to be occurring at same 0 time in S frame of reference unless that event occurs at origin. Simultaneity is relative that is what we have been discussing so this is one of the very interesting consequence of Lorentz transformation that one notices. The third thing is about the inverse transformation we have often talked about the inverse transformation it means the information is given or the coordinates of an event is given in S frame of reference and I want to find out these transformation or these particular coordinates in S frame of reference and we have said that there is a very simple prescription of that replace v by minus v change prime to 1 prime and unprime to prime you will get inverse transformation so this is what we have written as inverse transformation. Again we sort of insist on these things which are very obvious because as we will realize that you know the relativity tends to become little more complicated so it becomes somewhat easier to think that when I am applying direct transformation when I am applying inverse transformation that is all. So this is my inverse transformation which is X is equal to gamma X prime so remember I have changed the primes and instead of v I have put minus v so this becomes plus earlier there was a negative sign here plus vt prime y is equal to y prime z is equal to z prime t is equal to gamma t prime plus vx prime divided by c square. So this is the sign of this particular term has also changed because v has been replaced by minus v. Now let us come back to our old example that we have done in our last lecture in fact some form the other this particular example we have been discussing at various times. Now once we have obtained Lorentz transformation let us look at this particular transparency once more and this particular example once more. So let me remind you my old example this was S frame this is S prime frame of reference we have talked number of times about this S and S prime frame of reference what are the specialties about them y being parallel to y prime z be parallel to z prime relative motion along x direction origin o prime moving along x direction. A light was emitted from the origin at time t is equal to 0 at that time of course the origin of o prime was also coincident and this is moving in a particular direction this is small pulse assuming it to be highly localized so I can determine what is the location of this particular pulse at a given time and at a time 2 microsecond we had evaluated what will be the position of this particular pulse in S frame then in our last lecture using Galilean transformation we had found out the coordinate in x prime frame of reference found out the time then eventually evaluated the speed in s prime frame of reference and we found that this does not turn out to be c because Galilean transformation is not consistent with second postulate of special relativity. Now I am going to show that with this example with Lorentz transformation things would change and now observer in S prime would also notice the speed of that particular pulse to be c. So let us just quickly go through the problem once more. An observer in frame S sees a pulse of light emitted from origin at least equal to 0 which is moving with a speed of c in xy plane making an angle of tan inverse 3 by 4 with x axis. Find the position of the pulse in S at t is equal to 2 into 10 power minus 6 seconds which we call as 2 microsecond assuming it to be highly localized. This part of the problem there has not been any change. This transparency also there has not been any change because as far as the information in S is concerned it is all given there. I do not need any transformation equation so long all the information is given in my own frame I do not require Lorentz transformation. I do not require Galilean transformation it is a pure simple classical kinematics that is what we are doing here. So we have said that this will be the value of cos theta this will be value of sin theta. So ux will be equal to 0.8 c, ui will be equal to 0.6 c, uz will be equal to 0 exactly the same transparency which we have shown in our last lecture there is no change here. The change would occur when I go back to S frame of reference when I want to calculate these coordinates in S frame of reference. Let us first calculate the coordinates in S frame itself no transformation required no Lorentz transformation required. Just take simple kinematic equation x is equal to ux multiplied by t which is also exactly the same as my last transparency my transparency in the last lecture x is equal to ux multiplied by t which is 0.8 multiplied by this time which gives 480 meters y, ui multiplied by t which gives me 360 meter z is equal to 0 taking speed of light c as 3 into 10 power 8 meters per second. So as we had discussed last time according to an observer in S frame this the coordinates of this particular light pulse assuming it to be highly localized at 2 microsecond will be 480 meters 360 meters 0 x is equal to 480 y is equal to 360 z is equal to 0. Now I want to find out the coordinates of the same light pulse in S frame of reference I will need Lorentz transformation. So I would need x prime is equal to gamma multiplied by x minus vt remember earlier this gamma was 1 in Galilean transformation. So x prime would turn out to be different from x, x prime will not be equal to x I mean it will turn out to be different from whatever we have obtained last time because last time we evaluated this x prime as by using the equation equal to x minus vt y prime is nv equal to y z prime equal to z as per Lorentz transformation. So first I need to calculate gamma which I am calculating in the next transparency. If you remember gamma has been defined as 1 upon under root 1 minus v square by c square at a little later time we will discuss that normally relativity we do not expect the values of v to be larger than speed of light though this is generally a commonly accepted fact by now. Therefore the value of gamma that I expect will always be larger than 1 because this I am reducing something from 1. So if I take under root of this particular quantity this will be smaller than 1 and 1 divided by something which is smaller than 1 will always be larger than 1. This will be equal to 1 upon under root v was equal to 0.60 when I was going from s frame to s prime frame the relative speed was 0.6 times the speed of light. So this becomes 1 minus 0.6 square of you calculate 0.6 square this is 0.36 which 0.6 is a very nice number it gives you very nice value of gamma. So many examples we use this particular number. So this is 0.36 if you subtract 0.36 from 1 you will get 0.64 and if you take under root of 0.64 this becomes 0.8. So 1 divided by 0.8 becomes 1.25. So x prime now becomes 1.25 multiplied by 480 minus 0.6c multiplied by 2 microsecond in Galilean transformation this quantity was 1. Now this distance x prime turns out to be 150 meters y prime there is no change 360 meters z prime was equal to 0. It is anyway equal to 0. So now according to Lorentz transformation the coordinate of this particular light pulse will be x is equal to 150 meters y is equal to 360 meters z is equal to 0 which turns out to be different from what we have evaluated using Galilean transformation in our last lecture. Now not only these coordinates become different using Lorentz transformation even the time becomes different so that also has to be looked into. So I must also find out what is the time according to s prime observer. So let us assume that this observer in s frame made a measurement at time t is equal to 2 microsecond to find out where is the pulse of light. Let me call that as an event. Now according to an observer in s prime this event would occur at a different time and not at 2 microsecond. As far as emission of light pulse is concerned because this occurred at t is equal to 0. So t prime was also equal to 0 there was no issue. But now this event is no longer at the same p value as observed in s frame. So what will be the time of this event as seen in s prime frame of reference how do we find out use the fourth equation of Lorentz transformation. So this is what is the fourth equation of the Lorentz transformation which is t prime is equal to gamma t minus vx divided by c square. Gamma we had already calculated this is 1.25 so this one was 1.25. t was what was the t measured by s observer which is 2 microseconds. So this is 2 into 10 to the power minus 6 second. v is the relative velocity between the two frames which is 0.6 c so this becomes 0.6 c. X is the x coordinate of the event as seen in s frame which happens to be 480 meters. So this x must be substituting we must substitute for x 480 meters divided by c square taking c as 3 into 10 to the power 8 meters per second as we have normally been taking to make our numerical calculations easier. We must subtract this number multiply by 1.25 this particular event according to s prime observer is observed at 1.3 into 10 to the power minus 6 second. So what an observer in s frame would conclude? He would conclude that in a light pulse has been emitted from his origin remember as as per his observation is concerned it has also come from his origin because this event occurred at t is equal to 0 equal to t prime equal to 0 at the origins. But now the light travel light has travelled only for 1.3 to 10 to the power minus 6 second at a time when observer in s measured the position of the light pulse. So they will differ in their times and according to the observer in s prime the x coordinate will be 150 meters and y coordinate will be 360 meters. So according to him the light has travelled for a time 1.3 into 10 to the power minus 6 second not 2 into 10 to the power minus 6 second. And have gone in x direction a distance of 150 meters and in y direction a distance of 360 meters. So I can calculate what will be the speed of the light as will be determined by an observer in s prime. All you have to divide do is to divide x and y by their times. Remember we have to be consistent in the frame. This information must be in s prime's own frame. x prime is also in his frame, y prime is also in his frame, t prime has to be in his frame. Let me just read whatever I have written. When observer in s measured the coordinate of the light pulse in his RR frame of reference the time was 2 into 10 to the power minus 6 second in s. The observer in s prime would differ. According to s prime the observer in s made a measurement not at 2 into 10 to the power minus 6 second but at 1.3 into 10 to the power minus 6 second. In this frame in this particular time the light pulse actually moved a distance of 150 meters in x prime direction and 360 meters in y prime direction. This is the statement which I was trying to make saying that distances have become different the time also has become different. Now if the observer tries to calculate the speed of light different components of the speed of light according to him I have put a prime here because this is an observation which has been made according to observer in s prime frame of reference. So this is u x prime would be the x distance traveled by the light which is 150 meters in a time of 1.3 into 10 to the power minus 6 second. This would give the x component of the speed of light or speed of that particular pulse which now turns out to be 1.154 into 10 to the power 8 meters per second. u y prime is 360 divided by the time. So it has traveled 360 meters of distance along the y direction in a time of 1.3 into 10 to the power minus 6 second. So according to him the speed will be 360 divided by 1.3 into 10 to the power minus 6. This is 2.769 into 10 to the power 8 meters per second approximately. Of course u z prime is equal to 0. Now if everything is what I am saying is consistent then if I take this square plus this square plus this square and take under root of that I must get speed of light because according to observer in s prime frame also the this light pulse must travel with the speed of light that is the postulate with which we have started that is the postulate with which we had evolved Lorentz transformation. So we must get back the speed of light. This is what I have done here. The speed of this pulse in s prime now will be given as u prime is equal to under root u x prime square plus u y prime square plus u z prime square and if you just take the exact numbers you will get equal to 3 into 10 to the power 8 meters per second as expected that turns out to be the speed of light. Now you remember see last time what was the difference when we had evolved everything s per Galilean transformation. The coordinates had changed last time but this time was same. So this u x prime was turning out to be smaller than u x while u y prime was same as u y. So obviously when I was taking square and adding to evaluate the magnitude of the speeds this was turning out to be different. In fact at that time I had even commented that if I want to make the speed of the light same in the two frames probably x prime has to be different from what we have evaluated and also probably t prime also has to be different. Now we know if we believe in Lorentz transformation which we today believe that the x prime has also changed from Galilean value t prime also has changed from Galilean value and thus making eventually the speed of light same in both the frames. So let me read here again we have seen we the see that as per Lorentz transformation both x prime and t prime have changed s prime from what we had obtained earlier classically x prime has gone up by t prime has gone down making both the velocity components increase from classical values even u x prime has changed increased u y prime has also increased thus maintaining the speed of light c to be same both s and s frame of reference. Let me do a summary of whatever we have discussed today first thing that we did is we arrived at Lorentz transformation the well known Lorentz transformation which is consistent with the second postulate. So that is my first thing we finally arrived at Lorentz transformation which satisfies the second postulate. Then we discuss the second old the old example of light emission to show that indeed the speed of light is maintained in both the frames. Thank you.