 Let us begin with what we discussed yesterday's lecture. We discussed that the consistency of electromagnetic theory and relative velocity expression required the presence of a special frame which was termed as ether. So, this is what we discussed in detail, the concept of ether which if both the things have to exist coexist together the electromagnetic theory and the old traditional relative velocity formula then ether was a convenient way of no having both them live together. However experiments did not provide support for the concept of ether that is what we discussed in our last lecture. Then we talked about the postulates of special theory of relativity. These were the postulates of special theory of relativity that laws of physics are same in all inertial frames of reference and with lot of argument yesterday we had on the semantics I just removed it ether does not exist. Now, we realize that we now require new approach. The old traditional approach will not work especially the relative velocity addition formula by which we found out that velocity C plus V or C minus V or C plus U or C minus U is not going to work. So, probably we require a different formula which has to be consistent with the second postulate of special theory of relativity which says that speed of light should be same in all frames. So, this new formula is needed which can ensure that speed of light is same in all all inertial frames. Unfortunately that is not just the enough thing this formulation must also be consistent with the first postulate which says that laws of physics should be same in all the inertial frames. So, what we will discuss first is how the relative velocity formula has to be modified. Then we will discuss that that is not enough you require something more in order that this particular thing relative velocity formula becomes consistent or other conservation laws become consistent with the first postulate which is that all I mean all inertial frames of reference you can apply to all the physics laws. Now, let us first discuss something what we call as a concept of transformation I think all of us knows what is a transformation, but nevertheless it is not a bad idea to formalize it put into somewhat formal language. If we have been given some information about any dynamical quantity in a given frame of reference and if we need to know equivalent quantities in a different frame of reference in another frame of reference the set of equations that we are going to use are called transformation. I mean as I had told you yesterday with class you know my experience even with IIT students is that when they solve mechanics problem biggest confusion is about the frames of reference people tend to confuse between the frames ok. I mean standard question that you know you throw a ball in the train ok with reference to outside observer it goes and makes goes up and comes like that while with respect to inner person you know it goes up and comes down comes down and people will use the idea of one frame and mix it with the idea of the another frame. So, these are very very simple things but you know I mean one thing which we always advise when you people solve mechanics problem you should fix your frame of reference. Now question is that I have fixed one frame of reference, but because of certain other reason I am interested in finding out equivalent quantities in any other given frame of reference then I have to use certain equations to find out these particular quantities in another frame of reference. For example, a simple thing would be just I know the coordinates of a particular particle at a given time in a frame of reference and I want to find out equivalent coordinates at that particular time in another frame of reference or we find the velocity of a particular particle in a given frame of reference and I want to find out the velocity of that particular particle in a different frame of reference or I know the momentum in a particular frame of reference I want to find out the momentum of the particle in a different frame of reference. So, the equations that we use to relate these two are called transformation equations. So, they are very type of transformation. For example, the coordinate transformation, we are just looking at the transformation of coordinates. The velocity transformation which I just now told that if you know the velocity components in a frame of reference, remember velocity is a vector. So, we talk about the components. So, if you know the velocity components in a frame of reference, I want to find out the velocity components in a different frame of reference, then the equation that I use is a transformation equation. Now, let us first before we go to actual well-known Lorentz transformation which is relativistically correct. Let us first go to the classical transformation which we are all familiar in the classical physics. So, no relativity involved here. So, that we just understand the transformation and then translate it over to the Lorentz transformation which is the relativistically correct transformation. So, I am formulating the problem. This is important to realize that all the equations that we are going to write is valid under these conditions. These conditions appear to be somewhat special, but actual in a real sense they are not really special because I mean in principle physics will not change. If you change a set of axis or say it is a change of what you are calling the x axis start calling as a y axis or if you want to take erect x axis anywhere in the space physics will not change. So, why not choose a set of axis which are more convenient more comfortable to deal with and because relativity involves lot of severe mathematics. So, why not make our mathematics simple and use somewhat simple concept. So, what we say that we take two frames of reference, two inertial frame of reference which one we call as s, another we call as s dash or s prime as it is normally called and we choose x and x prime axis in such a fashion. I mean as I say there is one particular frame of reference. Now in principle I can choose whatever x axis I want. I have another frame of reference there also I can choose whatever x axis I want, but I decide to choose x axis such that the relative velocity is always in the x direction. So, this is what I am always assuming that the relative velocity between the two frames is always along the x direction. So, suppose there is a one frame which is moving I mean there is one particular frame here, another frame is moving like this ok. So, whatever is this particular direction I choose my axis along that particular direction. It just makes our life little simple as far as mathematics is concerned. We always assume that y and y prime axes are always parallel. It means y axis of s frame and y axis of s prime frame they are always parallel to each other. Similarly, the z axis of s frame is always parallel with the z prime axes. This is what I have drawn here. So, one is s frame which I am calling as a red frame another is s prime frame which I am calling as a blue frame just to make it little more attractive from the point of view of color ok. So, s frame has x axis y axis and z axis as you can see x and x prime axes are always coinciding with each other. y and y prime axes are parallel to each other, z and z parallel like prime axes are parallel to each other as these two frames departs from each other because there is a relative velocity between these two frames or come together ok. We assume that the velocity direction is always along the x direction. So, the velocity direction is always in this particular fashion and we are assuming it to be positive. So, it means at a later time this particular blue frame will move more towards the x axis of the s frame. So, these are the special choices of axes which we have taken which by no means is loses the generality because as I said x prime axes are always under our control. So, we just decide that if I know I am in taking a frame there is another frame which is moving with respect to me ok. I decide that whatever is the velocity direction I will call that as my x axis and physics is not going to change if I choose a different x axis. So, this is what I am saying there is another thing which we are assumption which we are making is that at a particular time the origin o and o prime of the two frames would have been coincident. We assume that time is measured from that particular point of view. So, time t is equal to 0 is when both the origins of the frames coincide. So, o and o prime coincided at time t is equal to 0 when both the frames were exactly coincident the both the axes were actually coincident. So, this what I said frame s prime moves with respect to s along x axis with a velocity v. Of course, this will have velocity v will have only the x component because we are assuming that is what we are assuming clock of each observer because there is observer sitting in s frame there is an observer which is sitting in s frame of reference and both these observer synchronize their watch they just put time t is equal to 0 when they pass at their origin clock of each observer is set to 0 when their origins coincide. So, when person sees that there is origin of other frame of reference at that time he starts his watch similarly that person when that person finds that another frame of reference moving related to him ok when it is his origin coincides with his origin he starts the stopwatch alright. So, this is the way what we are assuming as a special choice of axes this I will not derive this is very very simple thing this is what is called Galilean transformation which is essentially a coordinate transformation of the classical transformation which tells that the x coordinate of any observation see any event or any particular observation at a given time t will be given by x prime is equal to x minus v t because you can see very easily that this particular thing will be moving the distance at a particular time t distance between o and o prime will be v times t. So, as far as x coordinate is concerned that x coordinate will differ by v t alright suppose we have a particular point here if I measure x coordinate of this particular thing as coordinate in s frame of reference will be this while in s it will be like this. So, basically x prime will be equal to x minus v t while because y and y prime has to be same z and z prime have to be same. So, this is what is standard Galilean transformation and we call is inverse transformation it means the information that have been measured are measured in s prime frame of reference and we want to find out the information in s frame. So, we have s frame and we have s prime frame of reference if the observations were made in s frame and what we want to find out what an observer in s prime will make the observation that is given by what we call as the direct transformation if the observations were made in s frame of reference then we want to know what are the equivalent observation in s frame of reference then we will use what is called inverse transformation actually we relatively we always use the direct transformation and inverse transformation they are always related to each other in principle even if you give one this should be alright all that you are doing change prime to un prime un prime to prime and change the sign of v that is the way it will always be able to go from any direct transformation to an inverse transformation that is all it requires to do. Now, here there is one particular thing which is we have written p prime is equal to p this is something which normally in classical mechanics we never discuss and we have when we are taught Newton's law of motion we have never talked about time in fact it appears that Newton has mentioned about time in his principia he assumes that time something which is uniformly flowing across the system or something like that he makes a statement but generally we always in sort of ignore time we always feel that whatever is that time measured in a frame of reference for any particular event okay let us fix our ideas in relativity it is always important to talk in terms of events something happens okay if I make a measurement of the coordinate of a particle this is what is an event okay this events coordinates I know the same events coordinate event is event if it has if there is an observation which has been made this event will be observed by all the frames of reference okay now we find out the coordinates of that particular event in a given frame and for the same event we want to find out the coordinate in a different frame so we generally talk in the language of event which makes it comparatively simple otherwise ideas get a little bit scattered so in classical mechanics we assume that if an event has occurred if I have made an observation at a given time t then according to another observer that observation will also appear to be made at the same time t okay so we sort of implicitly assumed t prime is equal to t though we never talked about it and this is the thing which Einstein objected this particular last statement that t prime is equal to t is the one statement if you objected how we will see now it is easy to derive what we call as a classical velocity transformation you just differentiate with respect to time these coordinates I will not go into the details as we said we will not go into the mathematics and you get a simple straight way what we call as a relative velocity formula this is nothing but relative velocity formula that u x prime will be equal to ux minus v so x coordinate of velocity will differ in the two frame of reference by v the y component of the velocity and z component of the velocity will remain same remember all these transformations are valid only with respect to that special choice of axes that we have taken that is something which have to be kept in mind because we have defined our axes set of axes we have defined how our time is going to be measured how our coordinates are going to be measured with reference to that particular sets only these equations are correct so it should be very very clear so this is what we call as a velocity transformation formula this is the same as the old relative velocity formula which is inconsistent with the second postulate as we have just now seen because if this formula is valid has to be valid for light then in that case it is not going to be consistent with the second postulate because then we will find that the velocity of light will be different in different frames and that is what we want to avoid that is what we do not want we want that velocity of light should be same in all the frame if we believe in the second postulate of Einstein and this is the inverse velocity formula let us not go into the detail of the inverse velocity formula so what we have to do we have to now look for a transformation which is consistent with the second postulate and can make speed of light same in all inertial frames that is what we are aiming at and Einstein after a lot of thought in fact that is what he said after a lot of thought he thought it is basically time which is the culprit basically the assumption that t prime is equal to t that both the observers make the observation at the same for an event both the observers will claim the time to be same that is what he objected okay so let us give some examples try to understand how it is possible this is what I said time is suspect now let us like try to take some sort of examples generally time is measured the way we measure time is sort of related to simultaneity of two events if I say that a train starts from CST station let us say at 10 am it essentially means that when the train starts there is one event the train has left the station just left the station and there is another event which is the time in my watch I would look at the watch and I find it is exactly 10 o'clock okay and these two events are simultaneous it means when my watch shows 10 o'clock that is one event and when the train departs that is the second event both have occurred at the same time okay similarly I can say my class will start at 1130 it means when this particular time the watch will be 1130 and when my class starts these two events will be together they have occurred simultaneously they will happen appear to be in the same time so this is what I say it essentially relates to some sort of simultaneity so a train leaving at 10 am implies the train leaving and talk during 10 am are simultaneous events we now want to show that if second postulate is correct if Einstein second postulate is correct then simultaneity is relative two events which appear to be simultaneous in a given frame may not appear to be simultaneous in a different frame so let us first look at that and convince ourselves that time could be frame dependent then we will look into the actual transformation which is the Lorez transformation all right so let us take another example so let us assume that we are in a this is sort of a train compartment okay and I am sitting not standing just at the middle of this compartment then I throw a ball to the right hand side and a ball at the same time to the left hand side all right let us assume that this particular train is moving I am in a compartment this is moving relative to a particular velocity in let us say another frame of reference which let us suppose earth frame of reference let us assume that we are in a room now we describe two events event number one when this particular ball which I have thrown here hits the back wall let us suppose we are moving in this particular direction relative to let us say outside so let us call one event when this particular ball hits the back wall and even number two the second event when this particular ball hits the front wall so I have two balls in my hand I throw just like this okay this ball will go and hit this particular ball this particular ball will go and hit this particular ball let us assume that my velocity is enough so that the ball will not fall and hit the ground okay before it collides to the ball so this is what I have written an observer is exactly halfway in a running compartment of length l so let us assume that this length is l he throws two balls okay at the same time t prime is equal to zero so let us assume that t prime was equal to zero with the same speed to prime of course I must add that the velocity of the ball relative to me I throw with the exactly the same velocity as measured by him one towards the front wall and another towards the back wall is this idea clear so this is what I have drawn it's a sort of a card room of this particular thing that you know of course my man looks too big in comparison to compartment but it's just for illustration purposes that you know there is a ball here there is another ball so this particular person remember these velocities u prime because I have put primes okay are all measured related to s prime all right so that's the notation which should be very clear so that's what I have written u prime so I am sitting in s prime frame of reference and therefore the velocities with which I am throwing the balls okay are measured according to me okay so relative to me this particular ball and this particular ball moves with the same velocity which is u prime so this is thrown like this this is thrown like this so I describe my event event number one the first ball reaches the front ball okay my event number one is that this ball comes and hits this particular ball my event number two is that second ball comes and hits at the back wall now my question is very simple question are events one and two simultaneous okay implying whether they occur at the same time all right at the moment I am observer which is sitting s prime frame of reference okay you will realize that this particular ball has to travel a distance of l by 2 because this length of this compartment is l remember we are talking still of classical mechanics all right so this particular ball has to travel a distance of l by 2 because I am sitting half the way this ball also has to travel a distance of l by 2 this is thrown with a speed of u prime this is also thrown with a speed of u prime so time that it will take to travel from here to that particular wall will be l upon 2 divided by u time taken for this ball to reach this particular wall will be same l by 2 divided by u u prime all right so according to me I will find that this event of this particular ball hitting and this particular event of this particular ball hitting this ball are simultaneous they occur at the same time all right before I go to slight mathematics let us just first understand observation now let us imagine with respect to the observer on the ground all right now look at this particular person now what will happen according to this person when this ball has been thrown in this particular direction it is not actually thrown with a velocity u prime okay classical velocity Anishin formula he will find that actually the velocity of the ball is u prime plus v all right on the other hand according to this observer this ball does not travel with velocity u prime but travels with a velocity v minus u prime all right so according to this particular person this ball is actually traveling with a larger velocity while this particular ball is traveling with a smaller velocity okay remember the observation is now being made with respect to the observer in ground all right but on the other hand as this ball I am now looking with respect to this particular person as this particular ball is moving towards right this train compartment is also moving towards right all right so actually ball has to travel a larger distance to hit the ball to catch the ball all right on the other hand as far as this particular ball is concerned though it is moving with a slower velocity but this particular ball here is coming towards the ball so this particular ball has to travel a smaller distance to hit the ball okay and I can do a small calculation okay you will find that according to this particular observer also these two events will turn out to be simultaneous all right the classical mechanics has been designed in that particular fashion okay so this ball this ball travels with a larger velocity but travels also a larger distance this particular ball travels with a smaller velocity but travels a smaller distance and if you do a small calculation you will find that time taken for this particular ball to hit this ball will be exactly same as the time taken for this particular ball to hit this particular ball so let us first talk in s prime frame of reference that is the train frame of reference that we agreed that time for the first event was l by 2 u prime we just just now agreed that was for event number one that is the ball hitting the front wall for the second event the time was l divided by 2 u prime exactly the same thing nothing has changed as far as this particular person is concerned hence the two events are simultaneous in this frame of reference now if I go with respect to the ground observer the speed of the first ball using the inverse velocity transformation which I have used just now is u x is equal to u x prime plus v so the first ball travels with a larger speed u x prime plus v while the second ball travels with a smaller speed which is v minus u prime all right now we go to slightly different thing how much distance this particular ball will travel in the time t 1 if the time t 1 is the time taken for the collision so this will be v t 1 so first ball has to travel a distance of l by 2 plus v t 1 okay and it has to cover that distance with a speed u prime plus v the second ball has to travel a distance smaller by v t 2 okay and it has to travel the distance by v minus u prime you calculate t 1 prime and t 2 prime this will give you exactly the same thing okay in fact the t 1 is equal to t 2 is equal to t 1 prime is equal to t 2 prime all everyone will agree as as as well as prime will agree that both the events were simultaneous and those events occurred at a time l upon 2 u prime as far as observer s is concerned as far as observer s prime is concerned both will agree that this particular observer these particular events occurred at the same time and at a time l upon 2 u prime this was a purely a classical treatment now let us introduce light let us introduce the second postulate of special theory of relativity okay now instead of these balls let us assume that I have two small bulbs two torches I flash like this you know I flash like one in this side I flash on this at the same time so I have two similar type of torches I or laser beam or whatever you want to call it you flash exactly at the same time everything is identical except that instead of ball I am talking of now light okay and now I am assuming that second postulate of Einstein is correct that velocity of light is same in all the frames so that is what I write now imagine that the observer shines light instead of throwing balls which travel both in front and the back direction event one light reaches the front wall exactly the same thing instead of ball I have replaced it by light so my event one is that light reaches that wall my event two is that light reaches the back wall let us try to analyze this particular thing in s prime frame of reference this is the cartoon again for that instead of ball I have put some flashes of light now you will agree as that s whereas prime is concerned it makes no difference instead of u prime all that he has to do is to use c okay so time for the first event will be t1 prime is equal to l upon 2c time for the second event will be l upon 2c all right because all that has become different is that instead of that ball velocity which was u prime okay now it has become c all right now let us look from the point of view and observe on the ground now according to this particular observer if the second postulate is correct this velocity has not changed that particular person will also observe the velocity to be seen if second postulate is correct this velocity will also be seen if second postulate is correct but on the other hand according to this observer here this light has to travel a larger distance because this particular wall is actually traveling forward all right while this particular light has to travel a smaller distance because this particular wall is traveling towards the light you agree with that particular statement yeah if you could be anything i don t care because if it is small value then there should be hardly any effect i know see things that whether it is miserable or not that s a different question okay please you know see questions are always welcome see let let s not i mean there is no teaching without questions i mean that s first thing which you always say that all questions are all welcome there is no question which is a foolish question there is always a question i mean it could be different of perception see a thing is that v is very smaller if v turns out to be smaller very small in comparison to see the effect may be not so much miserable that s a different question okay i m i m not talking of at the moment experimental way of doing it i m all i m saying is that will there be a difference whether it s miserable or not that s a different question see theoretically speaking there will be a difference whether v is smaller whether v is larger okay does not make a difference there has to be a difference because basically in our classical mechanics things were saved the time turned out to be identical because of course one particular ball has to travel a larger distance but the velocity was also larger okay another ball has to travel a smaller distance but the velocity was also smaller so the two event which you are appeared to be simultaneous in s prime also turned out to be simultaneous in s frame basic difference is that instead of light if i am instead of ball if i am using light okay and if second postulate is correct okay first statement is no longer correct second statement is still correct this particular ball is still moving with respect to s prime frame but this particular light velocity has not changed so this particular light has to travel now a larger distance to reach but with the same velocity see while a light which has moved in this particular direction according to observer sitting on the ground has to move a smaller distance okay but with the same velocity see obviously according to the observer on the ground if second postulate is correct the second event will occur first and event number one will occur later because two velocities are identical but one has to travel a larger distance another has to travel a smaller distance so two events which appear to be simultaneous in s prime frame will not appear to be simultaneous in s frame the time difference may be very small that is a different question but they are different that is what is the idea okay it means the two events which are simultaneous in a frame may not appear to be simultaneous in any other frame it means the concept of time which was basically involving some simultaneity probably may not be correct it means the time that we are measuring may also be different in different frames it means t prime is equal to t or t is equal to t prime that we have written in the Galilean transformation may not be correct that is what Einstein started looking into it and sort of working on it yes so that means when the light gets reflected to the world and again come back to the person then they will not reach simultaneously one ray of light will come first and then the other ray of light suppose sir I design a thought experiment where there is a photo catalytic reaction chemical reaction for the person who is sitting inside the train for him the two lights are coming together so the reaction takes place for the other person who is standing outside the train for him one light comes faster and another light reaches later so for him the reaction doesn't takes place same chemical reaction so how will you interpret it no no no that's a I'm not very sure what you mean by chemical reaction because as far as I am concerned chemical reaction these two are sort of independent thing this particular chemical reaction this particular chemical reaction are independent so in a particular frame the chemical react two chemical reactions will appear to be occurring at the same time according to another person they will not appear to be occurring at the same time suppose sir there is a significant amount of light needed for a say photoelectric effect not chemical reaction okay let's say photoelectric effect okay see I sent a photon here a certain amount of electrons are started coming there okay now according to me electron coming from this particular source and this particular source they started together according to observer around they started to get different time so when the electrons started getting emitted from here first and electrons started getting emitted from there later okay if it is at the two ends but suppose light is coming back towards the person and meeting him again yeah then for the person who is standing on the train moving with the train for him the two rays of light are coming together meeting together yes so more electricity will be generated in the photoelectric effect let's discuss it out later I'm not sure what is what you have in mind see thing is that basically we have to talk in terms of event what we are talking of the events okay we should define our events let's try to find out what is the time of the event okay because when the light has to come the things will become different because then that particular light will change the role now this particular light has to travel a smaller distance okay now this light has to travel a larger distance the roles have been interchanged so things may become different so that I'm not sure whether I have answered your question but you know things that you have to look at the entire event holistically and talk in terms of events all right now I'm not calculating the time difference of the two events as seen in this these two frames because until we have done Lorentz transformation which is little more complex I don't want to go my idea is only to tell that the events which appear to be simultaneous in a given frame may not be simultaneous in another frame that was one example which I just now I have a question yeah actually you are suggesting to teach was relativity in simultaneous yeah before before you deal with the Lorentz transformation that's what I like to do yeah but if some student who I mean who knows about Lorentz transformation I mean at least contraction he doesn't know about the expression and all these things when he has a hard of contraction he may ask that the length will be contracted and how are you see the thing is that you know it does happen and I do also happen so there are a lot of students who are already well read and they have a lot of I mean in fact one of my colleagues I mean let me be honest I'm not a specialist in specialty relativity I do not work in the area of relativity I'm not a theoretical physicist either but some of the people who are in my department who work on relativity I mean first thing which they always say that you must tell to the students to avoid use of length contraction as far as possible because length contraction is most of the time misuse rather than be used okay I will not be able to talk about length contraction in this particular course because we don't have that much time okay but if you are interested we can discuss it later see length contraction always involves what you call it a proper length and when this particular proper length is to be measured in a different frame of reference unfortunately what happens whenever a student sees a length first thing they will do contract it okay so I want to totally avoid the use of length contraction in fact these are shortcuts length contraction time dilation are shortcut methods of solving the problems and many times people get tend to misuse them I say always apply Lorentz transformation directly do not talk about length contraction okay if length contraction comes it will come automatically okay unless you are very very sure at least talk of the two events talk what happens whether the length that you are talking is really a proper length okay when you are talking of proper length proper length by the way is the length which is measured in a frame of reference in which the rod is at rest all right so then only we talk about it yes sir I would like to add a point in the Lorentz transformation derivation that we have there itself the concept of simultaneity is hidden implicit because we calculate the a point where a light has reached and then we calculate the distances that point is actually observed at two different times by the two observers so simultaneity is actually hidden in that derivation it's automatically part of there yes what would happen if that's a value value of v becomes equal to c that's a very interesting question because you know see unfortunately at when you are talking of velocity v equal to c or v greater than c many strange things start happening at v is equal to c you start having a discontinuity in the medium I mean all the expression when v becomes greater than c then all the expressions start becoming imaginary I mean mathematically people that's the argument people will give that velocity is greater than velocity of light should not be varied but there are better reasons in fact again I will not be able to talk about it if you want you can look my lectures which I have given an NPTEL okay the postulate tells us that no no no no see postulate does not talk about the speed of light postulate only talks the speed of light being constant yes now when we are talking of this particular thing that why velocity is getting sorry sir yeah actually we I mean speed of light is equal to c with respect to any observer okay any of any inertial observer so no observer can move I mean no observer can be a co-mover of light see I'm not very sure whether this is really implied in that statement all that's saying that it's if someone's speed is c with respect to me no no see I have to look at my relative velocity formula the things that I have not yet looked at my relative velocity formula if I look at my relative velocity formula what happens at c is equal to c that v is equal to c that becomes a little more complex but the things that even if I'm moving with the speed of 0.999 c this way another observer is moving with the speed of 0.999 c if you go to another frame of reference you'll still find a relative velocity smaller than c okay but if you ask you tend close to c this will also tend to c so tend to c but what I wanted to say is the better argument which I give in terms of why speeds greater than the speed of light should not be valid allowed is the more physical argument which is we call as a causality that the cause must precede the outcome see as I generally say that if I shoot somebody it should not happen in some other frame of reference that that person dies first before the gun is shot so it means see when there are two events which are related okay one event you know see I mean two events could be totally unrelated for example this train is starting this class starting and a train is starting from vd station okay these are totally uncorrelated statement relation okay they can always differ in a one particular frame of reference it may happen the class started first and another frame of reference the train started first that that may be alright but if that train moved because of my class is starting which of course not doesn't happen we can always imagine two events like for example somebody dying because somebody shooting okay that person died because somebody shot it okay then these are cause and outcome related so second event happened because the first event happened okay in that case it should not be possible that in any other frame of reference the cause appears to be later and the outcome appears to be first that you first see in the particular frame of reference that person has died then we find out that the gun has been shot okay so that should not happen and that's only possible if we avoid velocities greater than velocity flight so this is what is called causality the concept of causality and this is the causality which I think is a much more important physical ground for saying that velocity is greater than velocity of light are not acceptable but as you see I agree with this particular thing I normally try because as I say I find students still being very shocked when I say that t is not equal to t prime in fact almost every year somebody some of the student feels very disturbed and will always ask the question is not there isn't there any other way so this is the way unfortunately I mean this is a really shocking statement so therefore I want to emphasize on this statement first emphasize that t and t prime can be different then only talk about Lorentz transformation rather than talk about Lorentz transformation first and then telling that you know t is different from t prime yeah see in Lorentz transformation you get always an imaginary number see see the thing is that that's a more of a mathematical treatment you know I mean it's true that you know see I mean let me just talk you know let me call it a sort of loud thinking you know see when there was a you know this our the famous experiment you know which was being happened you know LHC experiment was being conducted there was once a report that particles greater than particles speed of light has probably been seen of course that was denied after some time that this particular thing was not very very sure and then a lot of people were thinking even already one will you are thinking that no see can't affect you can't the rest must be imaginary because this particular particle can you know in principle can never be put to rest okay so I mean thing is that you know if you have that even probably people can get across I mean we use imaginary numbers for many things in I mean see a whole your whole of your quantum mechanics wave function based on the images but we don't measure imaginary numbers we don't measure imaginary numbers so things that see question is that when you can measure only real numbers but in order to achieve at real numbers okay you may always use imaginary numbers okay that's what we do in quantum mechanics for some time yeah so those are more measured for take no so I mean that's what I'm see that's what I'm saying it's a sort of loud thinking we're not thinking that you know rest mass you know for example the particle can never be good to rest so what is the rest mass we can can't be called as mass is imaginary can't we make all those numbers you know even real I mean see nobody has visualized those things so it's I mean it's probably not very good idea to talk about it later but things that maybe there is a way of handling those numbers maybe I do not know whether I mean I'm not saying that there will be a way okay maybe there is a way okay but the physics arguments we always believe much more strongly okay because if this would have if causality was violated someday I would have seen in my frame of reference some causality to be violated which I have never seen someday in my frame of reference I would have seen somewhere in sky that the cause occurred later and outcome came first I have never seen that thing but when we say the time is different yeah and we have to think about that then you have to think about what I don't know but they have to predict the future see I'm not very sure whether it means that because see things that you know let me also put it there are a lot of paradoxes there are a lot of things which people talk about especially eligibility I take a concentrated effort of not to talk to them talk about them because many of these things are looking very attractive and probably glamorous to talk about those things but eventually when you start dealing with physics you always find that there is a loophole in their argument okay so that's the reason I purposely avoid talking and I want to talk of mainly those things which are directly physical so none of the paradoxes twin paradoxes etc all those things looking at future all those things maybe I want to talk about it because they always assume certain type of things see you can always happen and then you can meet together and then you can have exchange notes now these are there are many ifs and buts in there so they are more of a you know sort of glamorous thing to talk about their relativity but I normally avoid those things you know as I said you know okay let's go ahead and okay so this is what I had said that in s prime frame these are the two events which are simultaneous now according to s prime the two events are simultaneous this is because s prime the light covers the same distance in front direction as back with the same speed now according to s observer which is the observer on ground the speed of light is still see in both the directions according to the second postulate but it has to travel a larger distance to reach the front pole than the bi-fall so the two events in s prime frame of reference would not turn out to be simultaneous therefore probably it's not a bad idea to assume that time is also a frame dependent quantity and that's what Einstein did so this is what is the conclusion of whatever we have discussed there is a simultaneity of two events in s prime but not in s simultaneity of two events thus depends on the frame chosen thus time may have to be taken as a frame dependent if we have to make c independent of frame that's what is the basic idea and talk little bit about that how time should be frame dependent quantity so I repeat as somebody has mentioned that I normally prefer to discuss this aspect first before discussing Lorentz transformation just because I think people are generally very uncomfortable with the idea of time being becoming frame dependent quantity so this probably I will go very fast this is the basic equations of course there are certain physical arguments by which we can arrive at these equations if you look at the Resnick's book they have given details of how these starting from the most general arguments how the transformation should come to this particular form but I will here start with this particular thing so I assume that instead of the Galilean transformation my transformation has become somewhat different basic difference here is that in Galilean transformation I had x prime is equal to x minus vt now this has been multiplied by a constant bxx prime bxx sorry these two things are concerned this is same as the Galilean transformation the third thing that I have made is time dependent on frame so t prime is no longer equal to t but t prime is related to x and t so t prime becomes btx into x so because this is related to x so I have put x symbol here this is related to t so I have to put the t symbol so btx into x plus btt into t now it is very clear that if my vxx was 1 and btx was equal to 0 and btt was also equal to 1 this will lead to a Galilean transformation so just we have generalized little more the Galilean transformation of course people can always ask the question why equation should be only in this particular form they are profine perform physics arguments on the basis of this particular thing as I said if you look at the Resnick's book he starts with the most general transformation he says let us suppose we are totally in dark so let us assume that x is now depends on x y z t x square y square z square x all those things then starts giving physics arguments in fact if you want to do that you know normally it takes about 1 1 1 2 hours itself to say how from those most general arguments I will arrive at these equations so all I want to say that there is this is not just randomly that I have chosen these things there is a physical argument on the basis of which you can argue it out that these equations should be of this particular form all that is now required to be known are these three constants bxx, btx and btt and the only thing that we have in our control is that I want that the speed of light should be same in all the inertial frame of reference which one here so this is btx this is btt from prime btx product of btx here no which which equation are you talking btx btx talking no here they see is that because this is a relation between x and time so this constant I am calling btx so dx is subscript so this I am calling as tt because this is relating time part of the component to time here because this is relating to x component actually this you can of course I mean in principle I should have written bxxx into bxt okay but on the other hand because both the constants turned out to be same which also we can argue it out therefore I have just used bxx okay if you are that that is what is your objection so this because there is x component here there is also a time component if you want I can avoid xx it does not matter so I mean it's just a matter of again the way you want to put a constant you can put just abcd and xyz you know it doesn't make any difference yeah I mean this one no that's what I say if you if any student asks about this question I would always say I mean either of you I do completely because one can put an argument on this particular thing see you can look at that origin the way the origins are displaced with respect to this this particular equation must be of the form x minus vt because see remember this particular origin has to travel related to another origin such that any event which happens here must translate x x minus an event which happens at origin here at x prime must happen at at the distance of vt with respect to x prime x prime. Why not xx y sir? I am sorry see what you are saying that why I take no I am not I mean that that that is what I am saying this constant you can call anything you know that does not matter no no no question is v11 and v14 same that is that is what I am trying to say see I want to put this particular equation always in the form of x minus vt that is what is the basic argument because if an event happens at the origin of s prime okay then according to s that event must be occurring at x is equal to vt because that is what is our condition that this particular frame of reference is moving related to the first frame of reference with a relative velocity of v. So if any event occur at x prime is equal to at origin of that thing that must occur at x is equal to vt with respect to the ground frame of reference. So therefore whatever you say if you put x is equal to vt x prime must be equal to 0 that is all okay therefore this equation has to be of this particular form there is no other way all right. So as I say for all these things there are arguments I mean I can start with the most general thing and come to this in fact even if you look at my video lectures where I have started with most general equation and really talked about all those things and come to this particular equation but that takes one full lecture only on that particular argument how this particular constant should be because there are a lot of physics argument actually they are very profound argument very interesting arguments okay but one can show that this equation must be of this particular form okay then our standard thing is that okay it is easy to see that this equation satisfy the condition of time being 0 when the origins of the two frame coincide because this is the way we started with that and now we take the standard derivation making velocity of light same in the two frame of reference. So I assume that at t is equal to t prime is equal to 0 that was the time when the origins of both the frames coincided a spherical light front was emitted from the origin okay so this particular frame of reference moves in a spherical wave front starting with the center at the origin now because of observer s as well as prime observer both will feel that velocity of light is same in all the directions so as the time keeps on progressing both of the frame of reference both the observers will feel that this spherical wave front is still moving with respect to their center with respect to their origin as center of this sphere. So this is what is the standard I mean every textbook gives this particular picture that though this particular wave front started when their origins were coincident but at a later time when their origins were separated even at that particular instant of time it would appear that this particular red observer will feel that this is a wave front with this origin as the center of the wave front this particular observer would also feel that there is a wave front which is spherical which is with the origin at his center o prime all right because according to both the observers light moves in the same with the same speed in all the directions it is independent of the relative velocity between the frames. I am not see remember I am not saying the source which emitted the light was fixed in which frame of reference I mean had it been a wall in a wall in classical mechanics okay I would have said whether a ball was thrown from the origin s or as origin of s or s prime but light is immaterial immaterial whether the origin was whether the light source was fixed in s or whether it was fixed in s prime or it was fixed neither in s or nor in s prime is immaterial okay. What is important that this particular light was emitted when the origins of the two frames are coincident where light source was fixed is of no consequence all right because whatever you do for light the light has to move with the same speed c in all the frames. Now once you do this particular thing you can always write that these are the equations of the wave fronts which are the sphere equations are spherical a sphere the radius at any given time t will be c times t where t is the time measured in that particular person's frame of reference. So remember when we write these particular equations we write x square plus y square plus z square c square t square because this is according to a software word all the observations are being made in that person's frame of reference here everything is being made in s prime's frame of reference. So here I am using x prime here I am using y prime z prime and here also I am using t prime okay remember we cannot get confused between the frames of reference we have to be consistent in our own frame of reference as far as s prime observer is concerned according to that person the coordinates are x prime y prime z prime time is t prime according to the person in s frame of reference coordinates are x y z and time is t okay once we have done that then you all substitute those equations these are things. Now c t prime is equal to I do not say anything where the c t prime is equal to anything see I am there are two observers which are drawn the same radius no that is what I am okay I have drawn so that radius appears to be same you can treat it to be different this is only a pictorial representation of the spherical wave front okay the radius could be different okay depending on at what you I mean then we are talking of the events so that that don't get guided or misguided by this particular thing okay so probably I should have drawn a slightly different idea see the origins were concerned at the time when the wave front was emitted okay at that time the radii was zero radius was zero okay now as for a time they keep on moving apart from each other okay but both of them will feel that the wave front is still spherical and their origin is still the center is still the center that's all I am trying to say all right because it does not matter c moves with the same speed in every direction that's all I am trying to say okay see when you want to compare what is happening between s and s prime it's a little more complex because you have to look at Lawrence transform you have to talk about event okay when you made the event and measurement and things like that you know that's a little more complex so let's not go into the okay so these are just the equations which I will not go into the details of this thing that finally this is what you land up is what you call as the Lawrence transformation normally we use this type of you know abbreviations we always call beta is equal to v upon c very very standard notation I mean anybody in relativity if you call beta everyone knows this is what we call as a boost factor v upon c and gamma is 1 upon under root 1 minus beta square so in a simplified form this is the way your Lawrence transformation looks like x prime is equal to gamma x minus y t instead of 1 it has become gamma and this is t prime is equal to gamma t minus v x upon c square this is the well known Lawrence transformation basically there is certain observation which I will not be able to discuss too much in detail but they are sort of obvious if you are having any doubt you can always discuss with me in the classical limit that when v is much smaller than c the Lawrence transformation reduces to Galilean transformation which can be very easily seen because gamma will be very close to 1 because gamma depends upon 1 upon under root 1 minus v square by c square so in fact you will start getting relativistic correction only so remember when v is equal to 0.1 c gamma will become 0.01 of that order so you will start getting approximately 1 percent correction when only your velocity becomes of the order of 0.1 c so so long your velocities are smaller than 0.1 c you are probably making hardly any difference by using classical equations only when your velocities are much larger then you will start getting effects relativistic effects and of course as many of we have just now discussed that if v is greater than c this gamma factor will become imaginary which we do not know as of now how to handle it the inverse transformation can be written just as I said same formula change the sign of v put prime to unprime unprime to prime now let us come to velocity transformation now here before we go to the velocity transformation let us be little clear about notation because this again one point where people get thoroughly confused so when I am talking of velocity transformation unfortunately there are three velocities which come into the picture so let us be very clear that we do not make a mistake about symbol v in relativity is always reserved between relative velocity of the frame okay and this v has to be constant because the two frames are inertial that we discussed right in our first lecture okay so as far as relative velocity between the frames is concerned is v now let us imagine that there is a particle which is moving this particular particle may not move with a constant velocity all right could have any arbitrary motion okay now when I am looking at that particles is instantaneous velocity velocity at a given time okay the symbol that i u is u so u is reserved for particle velocity v is reserved for relative velocity between frames let us be very very clear never get confused about this thing because as soon as three velocity expressions start coming people start getting little concerned about it v is reserved for relative velocity between the frames which has to be always constant symbol u is reserved for particle velocity if it is being observed in s frame of reference then that particular velocity will be called u if the velocity of the same particle is being observed in s frame of reference then I call it u prime we have three velocities u u prime v v relative velocity u velocity of the particle as observed in s frame u prime the velocity of the same particle being organized being observed in s frame of reference and of course u and u prime need not be constant because the particle could be accelerated see I might be looking at the dynamics of a particular particle I might have applied a force on that particular particle that particle might be accelerating I may be looking into the dynamics of that particular particle okay therefore there is no restriction on the particle velocity but on the relative velocity between the frames that has to be constant because the two frames are in Russia frame so v is relative velocity between frames constant as a function of time u instantaneous velocity of particle in s need not be constant u prime instantaneous velocity of the particle in s prime need not be constant now generally we talk about events unfortunately I had not been able to give some examples which I would have loved to give probably in December course I will add some examples which makes things little more clearer before we come to the velocity transformation especially length contraction and time dilation I have totally avoided which normally we try to do specifically to make students careful when to use length contraction when to use time dilation okay or better not to use them so let us suppose there are two events first in s frame we observe a particular particle and find make an observation about the location of a particular particle and that particular observation we make at time t1 so this becomes my event see I observe a particle at given time here so at that time my watch shows time t1 and I find that it is located at x is equal to x1 this is one observation then after a little bit of time my watch starts showing time t2 then again I make a second observation and find out what is the location of the particle and I find that particle now exists at x is equal to x2 these are two events observation was made about the location of the particle which happened at x is equal to x1 let us assume y and z equal to 0 we will introduce y and z all right and we made the second observation of the same particle in s frame of reference and I found out that now this particular particle is located at time at x2 and at that time I watch showed time t2 this is particle this is event 1 particle found at x1 at time t1 event 2 particle found at x2 at time at x2 at time t is equal to t2 even if the velocity of the particle is not constant if I take delta t st2 minus t1 in the limit delta t tending to 0 delta x upon delta t will be defined as the velocity of the particle measured in s frame let us be very very clear if I am talking of s frame velocity delta x has to be measured in s frame delta t also has to be measured in s frame I cannot take delta t from any other frame of reference okay delta t has to be in my frame of reference delta s has to be in my frame of reference in a given problem if delta t has not been given in my frame of reference I have to use a proper transformation to get delta t prime from another frame of reference to this particular frame of reference alright now similarly if the motion is in general in three-dimension I can define my u x u y u z the x y and z component of the velocities in s frame as limit delta t tending to 0 delta x divided by delta t delta y divided by delta t etc etc so this is the way how I would define the x y and z component of the velocities in s frame of reference okay using two events similarly I can define the velocity component in s frame of reference I look at the same two events okay event number one was locating the particle position now this event was also being observed in s frame that particular person will feel that this particular event actually did not occur at x is equal to x x 1 but occurred at x is equal to x 1 prime and now because time is also frame dependent quantity so that person will observe that this event actually occurred at t is equal to t 1 prime similarly event number two according to that particular person would be observed at x 2 prime and at time t 2 prime and when that person calculates the velocity of that particular particle that person has to use his delta x prime his delta t prime and divide that okay he cannot take delta t of s frame and divide by the two is to take delta t of s prime frame over reference and divide by two so this is what is most important that it has to be delta x prime divided by delta t prime as delta t prime tends to 0 basic message is that a person has to be consistent in his or her frame of reference when I am making measurements in this particular frame of reference I am measuring time in my frame of reference I am measuring position in my frame of reference that is what is most important thing to realize now I use Lorentz transformation what I call as a differential mode so it was x is equal to minus gamma x prime is equal to minus gamma x minus vt so I take x 1 and x 2 take subtract these two I have sort of missed two or three steps which we call Lorentz transformation a differential mode just put deltas on all the variables if we take two things subtract one of one out of the two okay then substitute in these equations we had delta x prime to by the way delta t prime so delta x prime I have substituted from this thing gamma delta x minus v delta t delta t prime I have substituted this thing this gamma and gamma will cancel out so this is what I get then just remove manipulate the terms I get the expression ux prime is equal to ux minus v divided by 1 minus v ux upon c square remember in the Galilean velocity transformation this particular factor was just this this denominator is extra which ensures that relative velocity will never exceed speed of light similarly delta v things similarly this thing and eventually this is what happens as the final velocity transformation the only thing which has to be remembered here that in the first one there is no gamma in the denominator in the second one there is a gamma denominator second third also there is a gamma denominator see it means u y prime is no longer equal to u i remember y was prime was equal to y but u y prime is not equal to u y similarly u z prime is not equal to u z okay this is getting still divided by this particular factor that is what is the velocity transformation this is the inverse velocity transformation so we can see from this equation that if u is less than c in a frame of reference it will be same in all other frames also if u is equal to c in a frame it would be same in other frames also it can be very easily seen you can take squares of these things and add requires a little bit of mathematics you can show that if u is equal to c in one frame u prime has to be equal to c so this is the way we had derived it so these equations are consistent with the second postulate of special fluidity so these are the equations of velocity transformation that normally we should use yes sir all the derivations that you have done you have used second postulate of fluidity sir where is the first postulate that is what i am coming to that is what i am precisely i am coming to it okay so i i have a question notations of difference is there any reason for that you see that is the standard rotation that we use for every differentiation see we start with deltas which we are assuming that difference is sort of finite and let that difference tend to zero but we agree with inertial frames so it doesn't matter it doesn't matter see even if this is inertial frame the particle is moving from here to here and in principle if the particle is accelerated the distance delta x if i am taking at this as t is equal to one second and i am taking at t is equal to two seconds okay this delta x will be different so i if i want to look at this tendin as velocity i must make delta t tendin to zero but you my question was uh you have avoided the symbol of d dx did you see i mean no see the okay i mean i have not avoided you can write d dx there is absolutely no issue it's only to simplify the issue because i wanted to talk i mean the basic idea which actually to my students i always want to talk tell that in relativity you must always talk of events that's the way i want to talk because you know if you do not talk about events you are likely to make mistakes so therefore when i want to go from coordinate transformation to velocity transformation or from Lorentz transformation to velocity transformation okay i want to talk in terms of events so therefore i have defined event one and event two and then take delta x and then let it tend to zero okay if i don't have to talk about this i can straight away write d dx dt i mean there is absolutely nothing wrong in that thing but only because i want to practice to my students that whenever you talk you talk on terms of events so how i'm going from distance to velocity okay you talk of two events measure the position okay then it becomes easy but then people will ask okay whether this instantaneous velocity was the same time or different time now t is also frame dependent quantity you are always caught okay so that's why i always say may talk in terms of an event measurement is done okay now what are the coordinates of that particular event you can talk of those coordinates being viewed in a different frame of reference then you know what is if t is this much here t is equal to t naught what will be time in s frame of reference then it becomes easy to talk so it's always convenient to talk in terms of the in events otherwise one gets confused in these things and last statement which i have written that you can also show from this particular law in transformation that if at all it happens that you become greater than c in a frame of reference you can show that in every frame of reference you will turn out to be greater than c so you will not be able to find out a frame of reference in which this particular particle will be able to cross the limit u is equal to c it will always be exceeding u is equal to c if at all it happens okay i mean first two statements we can always talk about it that if u prime is less than c it will always turn out to be less than c in any other frame of reference and if u is equal to c then it will turn out to be equal to c in all other frames now i come back to the second thing which as you asked is law in transformation enough have you solved over a problem of relativity question is that unfortunately it is not okay we still need to do something more in relativity because i have to show that this particular thing is consistent with the first postulate okay and the first postulate says that all laws of physics should be same in all the inertial frame of reference now i want to give an example very very standard and simple example which says that if conservation of momentum which we believe is a fundamental law of physics is valid in a frame of reference if all that we are doing is introducing Lorentz transformation and the velocity transformation then that momentum will not turn out to be conserved in any given frame of reference so therefore we have to redefine momentum and eventually redefine energy which eventually led to the famous expression is equal to mc square okay so first of all i discuss that there is a need to define redefine momentum so i take a simple example which is very well known example almost all students know about from the high school this is called a completely inelastic collision so i have further simplified it by taking masses to be same so there is a one particular frame of reference in which you find that there is a mass m which moves to the right hand side with the speed of 0.6 c there is another mass same mass m which moves to the left hand side with the speed of 0.6 c these two masses collide and the collision is completely inelastic that in classical mechanics means that these two particles get stuck to each other so finally you find that these particles are just stationary sitting together okay and you can see that in this particular process momentum will be conserved because the initial momentum will be m multiplied by 0.6 c and for this particular particle for this particular particle will be m into minus 0.6 c it means the sum of the momentum of the two particles will be 0 and when the two particles are at rest then momentum is 0 because the velocity is 0 okay in this process we always tell that energy is not conserved strictly speaking energy is always conserved it is not the mechanical it is the mechanical energy which is not get conserved okay the energy is always conserved in the form that it gets generated to heat or sound or whatever some other effects of the energies but the traditional mechanical energy is not conserved there are some collisions which are called elastic collisions okay which you differentiate in classical mechanics at our high school that okay these are the collisions in which both energy and momentum have to be conserved while collisions like this only momentum have to be conserved because energy is lost in something else so this is what it happens that initial momentum is equal to 0 and final momentum is equal to 0 momentum is conserved now let us view this particular collision from s prime so let us assume that I go to a frame of reference of this particular mass m there is another frame which is now erected with this mass m okay or let me put it like that this is a frame of reference in which this particular mass was initially at rest okay after collision it will not be at rest so let us assume that there is another frame of reference s prime which moves to the right hand side with the same speed 0.6c as that of mass m so it would appear that in this particular frame of reference this particular mass m is at rest okay and apply velocity transformation equation which we have just now derived and see whether we will find momentum to be conserved as we expect momentum will not appear to be conserved so let us view this collision from s prime frame which has v is equal to 0.6c using velocity transformation so this simple equation so ux minus v so I put ux minus v which is 0.6c because v is remember relative velocity of the frame which is equal to 0.6c and ux was also equal to 0.6c so numerator will become 0.6c minus 0.6c divided by 1 minus uxv upon c square there is no gamma in ux all right numerator is 0 so this particular velocity will be 0 which is expected because if I am moving in the same frame in a frame of reference which is the same velocity as that of particle then this particle will appear to be at rest and that particular frame of reference now let us look at the second velocity particle of the second velocity this particular particle actually in this particular frame moves in s frame in the other frame of reference was moving at velocity of minus 0.6c so if I have to find out this particular particle's velocity in s prime frame of reference then my ux will become minus 0.6c v was actually equal to 0.6c ux minus v minus 0.6c minus 0.6c divided by 1 minus u into vx upon c square okay there is one of them ux is negative so therefore it becomes plus and going little rapidly I am sorry so this particular velocity will turn out to be minus 1.2 divided by 1.36 u remember this velocity will always turn out to be smaller than c as we expect we had there will only be numerator then this will have been 1.2c which will have been larger than c but because of this denominator factor this velocity will turn out to be smaller than velocity of light now let us look at the composite particle which we have received which we have obtained in s frame of reference original frame of reference which was at rest so now according to s prime frame of reference what would be its velocity ux is equal to 0 v is 0.6c 1 minus u vx into c square u is becomes 0 minus 0.6c alright so according to this particular observer the two masses which have come to at rest their velocity will be minus 0.6c the initial momentum will be moment of the first particle was 0 because velocity was 0 I had calculated velocity as minus 1.2c upon 1.36 times fc this will be the velocity momentum final momentum will be this according to this observer the momentum has not been conserved so obviously this is not correct way of defining momentum I have to define momentum in a different way because otherwise this will be inconsistent with the first postulate of relativity either we say that conservation of momentum is not a good I mean it is not a fundamental law of physics if it is a fundamental law of physics then it must be obeyed in all the frame of reference and therefore we must redefine the momentum so without doing any four vectors I am just giving the definitions we define a quantity called rest mass this is a mass of a particle which is measured in a frame of reference in which the particle is at rest now we defined the momentum not as just m not times ux the way I have just now defined but we put a factor of gamma now you may again get confused because I have used gamma u earlier I have used gamma but remember I have told you right in the beginning there are three different velocities v u u prime all right when I use v to define gamma then I will call it gamma when I use u to define exactly similar expression I will call it gamma u this gamma u is u square by c square which is the particle velocity which is not the frame velocity this gamma u need not be constant because you need not be constant all right just a matter of nomenclature so gamma u has been defined as 1 upon under root 1 minus u square by c square all right the u will become different in different frame of reference it will become u prime and then the gamma corresponding gamma that I will define will be gamma u prime so like you have three different velocities you have three different gammas gamma corresponded to relative velocity we just call it gamma gamma u corresponding to the particle velocity in s frame that I will call gamma u gamma u need not be constant similarly particle velocity in s frame of reference I will call it gamma u prime all right that also need not be constant so this is the way I define if I know ux uy use it as the velocity component in a given frame as measured by me then using this ux uy use it I can find out what is u u square will be equal to ux square plus uy square plus uz square using this u I can calculate gamma u ok I have measured my velocities in my frame of reference I know ux uy use it I use this ux uy use it to find out u this u square is ux square plus uy square plus uz square ok once I know u square I find out my gamma u now instead of just multiplying m naught by ux I also multiply it by gamma u that is a new definition of momentum now similarly for the same particle if I go to s frame of reference that particular particle will experience will find out the speed of the particle as u prime and therefore if that particular person has to define the momentum in his or her frame of reference ok he would use these ux prime uy prime uz prime to calculate u prime square using u prime square he would calculate gamma u prime then take ux prime multiplied by m naught and also multiplied by gamma u prime that is the way we define the momentum in relativity now the second state which I make which again I am not proving here I mean requires a little bit more of time to do in classical mechanics as just now I said we define processes which are elastic and inelastic processes you know like elastic collision inelastic collision as to be more specific elastic collisions are those in which energy and momentum both are conserved and the example which I gave you was a totally inelastic process totally inelastic collision in which only momentum was conserved energy is not conserved unfortunately relativity does not distinguish between any such processes there is no process in which only momentum is conserved momentum and energy both have to be conserved that is what I said only momentum energy can momentum conservation does the new definition of momentum guarantee universality of conservation of momentum it is still not it is not enough it can be shown that if you want the conservation processes to be valid over all the frame of reference not only momentum but energy must also be conserved and what energy should be conserved was a new definition of energy not the classical energy so what I said it defined new form of momentum and a new form of energy and once you define momentum and energy in the new form then any process even completely inelastic collision both energy and momentum have to be conserved there is no process in which momentum and energy either momentum is conserved or energy is conserved okay every process both momentum and energy have to be conserved so the answer is no we need to conserve energy also in contrast with the classical mechanics and this is a new definition of energy which as all of you know I am almost the whole world knows those people who are not even physicists know Einstein is known as is equal to mc square it is a new definition of energy that you have to define if you define energy in this particular fashion which is also m naught gamma u c square so if you know the particle velocity in your frame of reference you can calculate gamma u take m naught multiplied by gamma u multiplied by c square this is the expression for energy this expression of energy has no classical counterpart it is a totally new concept of energy in fact you can show that if u turns out to be equal to 0 the energy will still be there which is m naught u square because if u is equal to 0 gamma u turns out to be equal to 1 and in that case even if the particle is at rest the particle will have energy which is called m naught c square which generally we call as the rest mass energy the kinetic energy was only the additional energy that you get in addition to the rest mass energy when the particle starts moving the new form of the energy does not resemble any classically known form of energy it is entirely new concept of energy energy gain or loss in different form would lead to an increase of mass this is something which is very very important say like in completely in last collision when the energy was lost okay in relativity energy was not lost actually it gained some other form of energy and that has effectively changed mass in fact mass is related to energy directly in fact I mean if you ask your particle faces they will always express the masses of the particle energy because energy and mass are linked by fundamental constant c square if you ask particle faces what is the mass of the electron you will say half m e v 0.51 m e v what is the mass of the you know proton 938 m e v see electron volt is the energy of the units of energy actually means that m naught c square is equal to 0.51 m e v mass can be expressed in the units of energy as the two are related to a fundamental constant kinetic energy is defined as m c square minus m naught c square you can show that this energy in the classical limit turns out to be equal to half m u square so basically these are our new definitions p is equal to m u e is equal to m c square kinetic energy is equal to m c square minus m naught c square and m is this expression m naught upon under root 1 minus u square by c square these are the new definitions which one has to take through of course this is the famous energy momentum relationship see like in any problem we have to convert energy into momentum this is the relationship e square is equal to p square c square plus m naught square c to the power 4 this is the relationship between kinetic energy and momentum which in fact can be shown that in classical limit is approximately equal to p square by 2 m i am sorry i am just hurrying to because these are just mathematics in fact these transparencies will be available to you so you can go through this is my last transparency in every process both energy and momentum have to be conserved in the complete inelastic collision discussed earlier the rest mass of the final particle will be larger than the sum of the rest masses of the initial particles of course there are certain cases in which there is no internal structure then for rest mass of electron will always be rest mass of electron it can never increase so energy and momentum both are not conserved that process will not be possible i'll give you example of such process in our later lectures i think just finish in time thank you yeah what do you had questions somebody said yes sir i am seeing that this is the last session on the relativity i have two questions about this i have two two questions yeah one question which is not included in this presentation is on twin paradox okay pardon twin paradox see as i am avoiding paradoxes at all yeah basically i'm confused in this so i i'm not ready to vote and i do i also don't want to read because things that you know i mean i can give you references or books where they will talk about paradoxes there is a book by professor Venkatraman who has been written which is essentially written on the paradoxes and relativity because they are basically wants why they are two observations different or same as measured by as i said most of these paradoxes always have something which is sort of hidden there okay because i mean which sort of avoids the use of kind of equation so i as i suggest that you know i can suggest you book you know this is a one book by professor Venkatraman who is written which essentially talks about the paradoxes and relativity okay so you can go with that book you know they will describe on all the second question is that sir what is the derivation of rest mass energy m0c square where from this relation comes see this is basically an assumption you know let me like that because when we find the totally let me first paraphrase this particular thing if you want to know why this i mean see there are two questions which i have not answered which can be answered if you look for example either it's a book or you look at my detail lectures which i have given you a reference from in the NPTEL see using the concept of four vectors okay it can be shown that if energy is put in this particular form and if momentum is put in this form this will guarantee the universality of conservation laws okay this particular thing which actually would have taken another three four lectures i am totally avoiding it okay using the concept of four masses you know you can always define certain quantities which are universal which will be valid in all the frame of references okay now if you use those particular concepts you can show that if momentum is defined in this way and energy is defined in this particular way then conservation laws will be universal okay now you can always argue the questions i mean one can always argue saying that why not this particular all i have said that if it is in this form it will obey but i have not said why cannot be any other form which will also be obeyed for this which i don't have an answer okay only thing i can say i mean remember any theory any postulate that we do has to be eventually experimentally verifiable nowadays we believe in these expressions because with all lot of those mass defects that we do in nuclear physics we are able to predict binding energy is very correctly we are able to get fission energy very very accurately okay therefore we believe in these laws if these would not have been experimentally verifiable at any extent okay people may not believe it so i am not showing why energy should be written only in this form and not any other form all i am writing is that if i write in this particular which also i by the way have not shown it if energy is written in this particular form and if momentum is written in this form okay universality of conservation of law will hold that's all i am writing yeah basically the thing is that when we derive the relation e equals to mc square to the students they should have asked when you are using m not c square already then what is the significance of deriving the e equals to mc square see i mean one thing which i always specifically mentioned that i am never deriving i never derived lawless law transformation okay i never derived for example schrodinger equation i never for example derived Newton's law of motion i never derived e is equal to mc square okay all these things have come from a set of arguments okay why i believe in the laws of motion because we have performed a large number of experiments that they seem to be obeyed at least in the classical limit so therefore we believe in the laws of motion okay relativity lawless transformation why do i believe it because now we have many experiments which can be explained only if the lawless transformation is valid therefore i believe it i have never derived so the of the word i am never deriving it i am just giving a set of arguments okay why this is the way it should be if it is done this way this will obey this property but you can always come out with an alternate way saying that no this is not correct if i define energy in something different way momentum in something different way is still conservative will hold good okay then where do i test whether this is correct or that correct only experiments right thank you yeah and if there are same masses then there is no resultant velocity many in many cases it is lost as heat and in other form sir if all the energy is lost as heat then how come the rest mass of the final particle be larger okay see when i am talking of these particular collisions at that time i have not assumed that loss of energy has taken place in any other better okay what i am saying that if there is no other loss then this particular rest mass will become this much now if it becomes cold okay then rest mass will go down if some certain amount of sound energy has gone down the rest mass will go down because all form of energy is contained in it is rest mass energy okay so if a particle has become hot its rest mass goes up the particle has become cool the rest mass comes down so it's just the variation of rest mass the only problem which is there which that's what i want to point out which probably i will discuss when i discuss Compton effect if these two particles happen to be electron then you cannot say that electron is a rest mass different from what we what is the standard particle standard electron see a bigger particle which can get heated then i can always talk that rest mass has gone up because it has become hot but if what happens to electrons if it becomes hot it will only start moving so there cannot be any other form of energy okay so in that particular cases if a conservation of energy and momentum is not violated that particular process will not take place for example one example which we always give that if there is an electron and if the photon is incident this electron can never absorb this photon so if free electron can never give you what you call as photoelectric effect if photon can never be absorbed by an electron because you cannot conserve both energy and momentum simultaneously okay so that process will not be possible no no it will not because energy not only energy momentum also has to be conserved say photon has a momentum of h nu upon c so that also should be conserved okay and you can show that if you make energy conserved momentum will go away so it is not possible the electron will never be able to absorb yeah so in fact either there i mean in photoelectric effect this electron is always bound only for bound electrons possible to see that till lorenz transformation length contraction time relation and velocity addition second postulate of relativity was only taken care of not the first postulate that's it that's it okay yeah okay thank you