 to the third part of our lecture on special relativity and having done the postulates of relativity and seen the consequences like length contraction. Let us move over to another interesting consequence that is called a time dilation. So just to set up the thing a little bit more, let us just have the results which we had in our last lecture, it is mainly on length contraction. So what did you see there? So what we saw there was if you have, if you have some object at rest in a certain frame, go on measure its length, find that the length of this rod, let us say it is L0. Now if you go on measure this in a different frame from which this rod appears to be moving with an uniform velocity, the length of the moving rod appears to be shortened in the direction of its motion. Of course, so that is the result that we got. So if you see the length of this object, this rod from S frame, remember that the rod is at rest in the S prime frame which is moving with a certain uniform velocity V. Then what is the length that you are going to see in the S frame? It is L0 times the root over of 1 minus V square by C square. So obviously L is less than L0. So the length of a moving rod or a moving object appears to be shortened as compared to the measured length in its rest frame. So well, so having seen something to do with the space coordinates, I mean that is the length contraction. What do we do? I mean what is it that we have with the time component and special relativity? Does something happen there? Remember in Galilean transformation of course, time in both these frames in S and S prime frames are the same. They are moving with a certain uniform velocity with each other. But is it so in special relativity? If you remember your Lorentz transformations, so there is a space component even in the time equation. And that in a sense leads to another interesting concept in consequence in special relativity. It is called time dilation. So what is it? What is the long and short of it? It is just moving clocks appear to run slow. So it is nothing to do with how good the machine inside the clock was. It is the concept of physics here. Or in other words the time expands for a moving body. So let us take an example, a short one. How do you measure such a thing? How do you get the feel of such a thing of things like time dilation? So if you consider a person who is inside a train and this train is moving with a certain velocity v with respect to the station of course, station or stations in this case. Then for the person who is sitting in this train, he matches his own watch. I mean he has synchronized at his watch with one of the station clocks earlier. Let us say then in the next station when the train passes, he looks at the station clock and then you will see that the times in his watch and the station clock will not match. So that is again what is meant by an example of a time dilation. But let us have a more pictorial example of this concept which will clarify it a little bit more. Say that you have two stations and then you have a synchronized clocks in these two stations. Synchronized in a sense that they give the same, the times are synchronized. So I should say that if you have two stations, you need two clocks there. So we have a two clocks, one on the left of your screen and one on the right of your screen as written as station A and station B and then there are clocks there. Then what happens? Suppose let us say a person moves with a certain velocity in a uniform velocity V in a train coach and then he has a watch on his wrist. Let us say so he is talking of a wrist watch and then he checks his time in this wrist watch and he adjusts it according to the time in station A so that he synchronizes his clock with station A. Now when he passes station B, what he will observe is that the time or whatever time it is in his wrist watch and whatever time it is in station B and then station clock they will not match. So this is a more pictorial way of saying the thing. Well, why is it so? Now for that we have to be a little bit more quantitative and so let us consider two events. I will explain what these events are. Let us just define what these space time coordinates of these events first. So we consider two events. We call it event 1 and event 2 seen from two different frames of reference S and S prime and as always we take the S prime frame that is moving with a certain uniform velocity with respect to S. If the space time coordinates of these two events in S frame that is x1, y1, z1 and t1 so that is measured at t1 and then the second event occurs at position x2, y2, z2 and at time t2 and correspondingly these two events occurs in the prime frame with x1 prime, y1 prime and z1 prime at this position and time t1 prime and then at the second event occurs at the positions x2 prime, y2 prime, z2 prime and at time t2 prime. Then what we have or if you want to be a little bit more specific now if you wish to define these events in terms of our train example we take the frame in which the so first we define what our S frame and this S prime frames are. So the frame in which the person with his clock at rest is the S prime frame. Now this frame so it is moving with a certain uniform velocity relative to station A and B and these are fixed in frame S. So now we define what our frames are so the stations are so you somebody is watching from the station so that is S frame and then the person in the train sitting in the train looking at his wrist watch so that is the S frame. So when the person passes station A let us call that as event 1 so it is got a certain position at certain time and similarly when the person passes station B let us call that as event 2 seen from two different frames of course the positions are different and the times are also different. So what we have remember in the S frame in which the person with his clock is at rest the events E1 and E2 has occurred at the same space coordinate but at different times why is that you see he is at rest within the train carriage let us say and then he just looks at his watch at the same position. So his position coordinates are the same but then he from his point of view he has looked at it at different times and that is why he gets different times. And just to keep the mathematics a little bit simpler we also take you know so moving in the common X and X prime direction so that the Y primes are also Y Ys and Zs are also the same here but in in terms of coordinates in the S frame we have X 1 prime is equal to X 2 prime and then Y 2 prime is equal to Y 1 prime and Z 2 prime is equal to Z 2 prime Z 1 prime is equal to Z 2 prime that is it but the times are different. So what are these times quote unquote times in S frame that is seen from the station so from the un prime frame if you call this the time T 1 so we just have the Lorentz contraction equations or Lorentz transformation equations to be more precise here. So T 1 that is T 1 prime plus V X prime by C square divided by root over of 1 minus V square by C square and T 2 is T 2 prime plus V X 2 by C square into root over of 1 minus V square by C square. So the time interval between these two events according to the observer in S so according to the measurement made in S frame so that is delta T let us say. So what is delta T now so delta T is just T 2 minus T 1 and that is T 2 prime minus T 1 prime plus V by C square into X 2 minus X 1 prime divided by the usual denominator root over of 1 minus V square by C square. Now what is the time interval according so that is the time interval of these two events according to the observer in S frame but in the S prime frame remember the positions of these two events were the same. So we have X 2 prime and X 1 prime are the same so X 2 prime minus X 1 prime is C O. So that will give us what that will give us that this delta T is delta T prime so we have defined delta T prime as the time difference as measured by the observer in the S frame. Remember the S frame is the one in which the person sitting in this train carriage is looking at his watch. So what we get so what we get is delta T prime is delta T into root over of 1 minus V square by C square. So just you get a feeling that you immediately see that delta T prime is actually less than this delta T here. Now what does it mean physically? It means that the time interval measured by the moving clock relative to S will be smaller than the time interval measured by the clock stationary in S. So in other words what we have is is confirmation of the statement with which we started with that moving clocks run slower. However I should remind you that if the velocity is much much less than the speed of light then you will immediately see that these two time intervals are the same. So they are approximately the same. So next time you come from let us say Delhi to Roorkee and then you look at the station clocks. Supposedly I mean you are moving with uniform velocity and all these things respect to the station in all cases. You should not you well you should not and you would not see this thing this time dilation. So having seen this thing let us have a look at what proper time is. So what is the proper frame? What is the proper frame of reference? So a reference frame in which body is at rest is called its proper frame we know that and we have also measured we know what a proper length is. So the length of the body measured in this proper frame is called its proper length. So the time measured by a clock at rest in the proper frame is the proper time. So just like the proper length is the proper time invariant well I think it is but let us just have a look at it once again. So the time interval so what is the proper time interval to be more precise? So it is the time interval measured between two events by a single clock at rest at the same place is the proper time interval between two events. We just had a example of a person sitting in a train looking at his watch at two different stations. So that is an example of you know position is not changing according to him in if he is sitting in the same place in the carriage so his position is not changing and then he looks at his own watch so that is a single clock and he is sitting at rest in the train so according to him he is measuring the proper time by a single clock. However I mean at the same time interval you know if you are going to measure it from another frame having a relative velocity with respect to the proper frame so you need two clocks at different places. So that is exactly what we had we had two stations and two clocks and then they had to be synchronized at the beginning well in a sense this time sometimes is also this kind of time is also time interval rather is also called non-proper. So if the velocity of the second frame be another frame be V relative to the proper frame with proper time interval delta tau between these events then the non-proper time interval delta T measured from the second frame of course by subtracting the time measured from two different clocks will be delta tau divided by root over a 1 minus V square by C square okay and so what is this delta tau that is the proper time so it is delta T into root over a 1 minus V square by C square okay. Now if in another frame in which it is moving with velocity V prime with respect to the proper frame if the time interval measured between the same two events is delta T prime then obviously delta T prime is again delta tau divided by root over now V prime squared by V squared so you immediately see that if we just write out the delta tau here the delta tau is nothing but delta T into root over of 1 minus V square by C square that is the time corresponding to this measured in the frame time interval measured in the frame it is moving with a certain velocity V then T prime this time interval measured in the frame in which it is moving with a certain speed V prime so that is equal to delta tau okay and so delta tau is invariant quantity okay so the proper time just like the proper length is an invariant quantity okay so let us look at an example okay or let us look at some examples here on these on these consequences of time dilation proper time and see what else we can do from here okay and let us take a simple but interesting example so you know in muon that is a subatomic particle okay and a muon is observed to move approximately something like 800 meters on an average during its lifetime in a laboratory okay so if one looks up the mean lifetime of a muon from one of these particle physics books or these data tables you see that it is something like 2 into 10 power minus 6 seconds okay so if we just go and write the velocity as the distance covered and then this mean lifetime so that is 800 meters by 2 into 10 power minus 6 seconds we end up with 4 into 10 to the power 8 meters per second okay this is much bigger than the speed of light okay now this is this is heresy in special relativity and what do we do I mean obviously we have not considered something so what is that well one needs to consider that what is what is given as mean lifetime of this muon is actually its proper lifetime okay so when you are measuring the length in the laboratory okay so you need to divide this by the this quote unquote improper lifetime so the measurement of the time and the measurement of the the distance this muon has traveled should be done in the same frame that is it okay so so this quote unquote improper lifetime of this muon as measured in a frame moving with this velocity v with respect to this proper frame so what is that so if we remember just a few slides back we talked of the proper time interval that is divided by root over 1 minus v square by c square so the proper time here is 2 into 10 power minus 6 seconds so divide that by root over 1 minus v square by c square okay and that is what we get the the speed of the muon is now 800 meters divided by delta t that is the improper lifetime and if you put in this this expression which we obtained earlier you will land land up with the speed of this muon to be a four fifth of the speed of flight so this is still less than the speed of flight okay and it doesn't you know it's not more than speed of flight as we had done as we had seen it would be if if you do it in a in a wrong sense of course okay so let me now move over to another interesting concept it's the relativity of simultaneity okay so and then I shall later on talk of things called earlier you know the concepts of earlier and later okay so the concepts like if you are if a certain event is you know following a sequence I mean some event has occurred earlier and then later and then does it appear to be in the same sequence in some other frame or not okay before that we do a little more basic thing it's called this relativity of simultaneity so so what's it that's going to that we're going to say here what we're saying is that two events are said to be simultaneous if they occur at the same instead of time of course I mean that's the that's the basic definition of what you would see for a simultaneous event okay but the question is if these events are to be simultaneous in one inertial frame so is it going to be simultaneous in another frame of reference which is you know so we can talk of inertial frame so we the frames here are moving with uniform velocity with respect to each other okay so if it is simultaneous in one frame is it going to be simultaneous in another frame okay so for that I think it'll be good if we see there are more pictorial fashion here so we talk of event one okay so we say event one so now it's it's it's occurred at a certain position so the x1, y1, z1 say at certain time in in s frame let's see okay and then the same event is seen to be at positions at at position x1 prime y1 prime z1 prime and then the time measured is t prime okay so that's the event seen from the the frames s and s prime so remember s prime is also moving with a certain uniform velocity with respect to the s frame okay now if we talk of another event okay um seen by the observer in s and s prime frame so it's occurred at a different position and time all together again so okay so what's it what's the relation between this between these time components so we know it we know it's t1 prime that's equal to t1 minus v x1 by c squared divided by root over 1 minus v square by c square okay so um and same thing for the for the second measured time in in the frame s prime so that's t2 prime is just t2 minus so minus v x2 by c square divided by 1 minus v square by c square remember this unprime things are for the s frame and then this prime things are for the s frame okay so what's the difference between these two times okay as seen well the difference between these two primes is obvious so it's t2 prime minus t1 prime so we find that's the same as t2 minus t1 so things on the right hand side are the the coordinates time that we have in the s frame okay so that's t2 minus t1 minus of v of x2 minus x1 by c square and divided by root over 1 minus v square by c square okay so if this events are simultaneous in in s frame okay then of course it's occurred you know at the same time so we take t1 is equal to t2 okay so that would mean that we have from this equation of difference of times in the prime frame so we have on the right hand side we have taken out t1 minus t2 okay however so the position so that's x1 minus x2 so that well it's not 0 why because if it's 0 so you wouldn't be able to distinguish between these events okay so so that's not 0 okay so that would mean that the difference of measured times of these two events in the prime frame is not 0 because none of the quantities here are 0 you can see here okay so what you get so what we get is the measured times in in the s frame need not be the same even if the times are the same in you know the measured times are the same in the s frame okay so in other words to be more precise what we have is that the events e1 and e2 okay although they may be simultaneous in one frame of reference they need not be simultaneous in another inertial frame okay so we come then to another interesting concept it's the this thing called the sequence of events the concept of earlier and later as we said I mean what we saw is that if something is happening you know earlier in a certain frame and then another event is happening later in a certain frame is this order of events preserved okay so why do we say that because we've just seen that you know this simultaneity itself is a is a relative concept here so it depends on I mean it need not be a something which is simultaneous in one frame of reference need not be a simultaneous in another frame of reference so will this occur here too I mean in this sequence of events so we consider two events e1 and e2 again and we see these events from two different frames s and s prime okay so this s prime frame it's moving with a certain uniform velocity with respect to s okay so we we've written down what the the space time coordinates of of these two events are in s frame and the s prime frame okay so the question that we now ask ourselves is that if these events occur in a certain sequence in one frame so can it occur in the reverse sequence in another frame so for that we again have to look at the the the difference of times as measured in the s and this the s prime frames okay so what's the difference of time measured in the s prime frame so that's t2 minus t1 that's the time measured in the difference of time measured in the s frame minus of the the velocity v into x2 minus x1 that's the those are the positions as measured in the s frame by c square then the usual factor root over of 1 minus v square in the denominator okay now if let's say event 2 has the second event has occurred later than event 1 in the frame s so obviously t2 minus t1 is positive okay now that means that for the reverse to happen in s frame okay what we have we have you know t2 minus t1 was positive in s frame but t2 prime minus t1 prime so that is not positive okay so in that case what it means is that we have the thing in the numerator so that's t2 minus t1 minus of v into x2 minus x1 divided by c square so that has to be negative okay that's obvious denominator here is positive okay so now if that is so we would require x2 minus x1 divided by t2 minus t1 to be greater than c square by v okay now now if these events are causally connected that is in a sense that event e2 is happening as a consequence of event 1 happening okay it's more like you know even a cricket field and then think of your favorite cricketer and then what he does is he autographs a ball and it throws it at you okay and so you catch the ball so he the cricketer your favorite cricketer autographing a ball and then that's called event 1 and then you catching you know the throw that he throws that event 2 so this has to be done at a certain you know so certain the message has to be passed at a certain speed here okay so you catching an autograph ball is in a sense dependent on the your cricketer signing the ball okay so unless the cricketer signs the ball and throws it to you you cannot catch the ball okay so in a sense some message has been passed in this case example of two events which are causally connected okay so in a sense what we have is that some message which is originating from the first event reaches the second the place the position of the second event traveling with some speed let's say some speed u okay then the minimum speed that this u should have is actually x2 minus x1 divided by t2 minus t1 difference of this positions and the difference of this measure that's the minimum thing that value that you should have okay u that is the the alphabet u I mean now what does it mean now this u in this case would have c square would be more than c square by v okay y because x2 minus x1 divided by t2 minus t1 that's more than c square by v so that's the condition one has to satisfy remember if the sequence of event is is reversed in in another frame and the events are causally connected okay but then we will end up with a problem the problem is that v here is less than c okay or at most it's c so at most it's c but so it cannot you cannot go more than speed of light but then what about u the alphabet u that is the speed u you see that it's more than c okay so this again is is not allowed so because why I said again because v is the speed s prime frame is moving uniformly with respect to frame s okay so in a sense it would mean that the message between the events e1 and e2 is traveling with a speed greater than the speed of light so this is not allowed okay so this condition is not fulfilled this condition is the wrong condition okay now since it's not possible to fulfill this condition if the events e1 and e2 are causally connected then it means that the concept of earlier and later between events will be preserved in all inertial frames okay so that's the point here so if the events are causally connected then it's going to be preserved but if the events are not causally connected okay so event second event let's see is happening independent of whether event one occurs or not okay in that case the opposite can be true okay so that's the case when they are not causally connected okay so what we saw is that all the this things called the simultaneity or the relativity of simultaneity tells us that if two events are simultaneous in certain inertial frame so it did not be simultaneous in the another inertial frame but the the concepts of earlier and later so that is preserved if the events that we are talking of they are causally connected okay so as usual let's have a few examples to clarify your concepts okay so let's take two events here e1 and e2 and in s frame and we give the position coordinates of these of these events here in the s frame and also write down what's the times here so here let's say e1 is happening at position x1 is x0 and then the you know the y and z are 0 here so okay and the x2 is happening at a position 2 x0 and then the the time t1 that's x0 by c so c being the speed of light and t2 is x0 by 2c okay now if the this event is simultaneous so see that in in s frame this this event is not simultaneous it's happening at different times okay but if this event is simultaneous in another frame so which is moving with a certain velocity v let's say along the common x x prime axis okay relative to the s frame what is this velocity and what is the time that what's the simultaneous time that's being measured in s frame okay so how do you go about doing these things so let's look at the time intervals as seen from the s frame and the s prime frame so since the events are simultaneous in the s prime frame so t2 minus t1 so that is 0 so if you look at the second step or the second equation so on this numerator you put in the times that we had for the event in for the events rather that we had in the s frame and then also the difference of the positions and divided by the root over of v square minus 1 minus v square by c square v being the velocity with which s prime frame is moving with respect to the s frame so we immediately get what we get the left inside to be 0 of course and we can immediately find what v is and you see that v is actually minus c by 2 okay so how do you find the time in the prime frame remember in the prime frame the events are simultaneous so so t1 prime that's nothing but t1 minus vx by c square divided by root over of 1 minus v square by c square and then what you're going to do is they're going to substitute t1 by the measured time you know for a certain position so if x is x0 we look this problem that the this was measured at time t1 is equal to x0 by c and we know the velocity now with which this prime frame is moving that's minus c by 2 so immediately get what this time is in the prime frame of reference so t1 prime is root 3 x0 by c okay so let's go to a another problem okay and well let's look at this problem first okay so in a certain inertial frame let's say so we have two events which occur at the same place and they're separated by a time interval of four seconds okay now in another inertial frame the question is what is the spatial separation what's the separation in this piece coordinates here between these two coordinates if these are separated by a time interval of six seconds okay so that's what we do we first try to find what's the velocity with which these frames are moving with respect to each other okay so call the prime frame that's the you know so the frame in which the time separation between these two events are six seconds so that's the prime frame say it call it like that so what's the corresponding you know portion of this time difference in the s frame how do you do it so it's t1 minus t2 or it's t2 minus t1 rather minus vx2 minus x1 divided by c square root over 1 minus v square by c square okay so here we have so you put instead of t2 minus t1 you put that to be 6 and t2 minus t1 you put that to be 4 and you find what v is and this v in this particular case would be over a 5 by 3 into the speed of light okay now having found that it's very easy to find what the spatial separation is so we just find what the spatial separation is so that's x2 prime minus x1 prime that's x2 minus x1 minus of v into t2 minus t1 into root over of 1 minus v square by c square we know the you know the spatial separations in the s frame that's 0 it's happened at the same place so what's the spatial separation in the prime frame so that actually is you put in the proper times here so I mean the proper times not in the sense of proper times proper length I mean you put in the right times here so that's t2 minus t1 and you get the difference to be minus 2 into root over of 5 c okay so the proper separation is the modulus of this of course so that's twice of root over of 5 c okay so we have seen so far it's another consequence of time dilation we have talked of you know we have talked how or why moving clocks run slow you know things are it's we derived it from we derived this consequence from the Lorentz transformations then we talked of the relativity of simultaneity so in which we showed that two events which are simultaneous in an inertial frame need not be simultaneous in another frame okay and then the third part we saw that the sequence of events they are preserved in all inertial frames if the events are if these events are causally connected okay so in the next lecture what you want to see is talk of Doppler effect in light and see what next we can do from there okay Doppler effect I think we are a bit familiar with Doppler effect in sound where you know if the the you have a source moving you know you're standing some you're standing at a distance and then from a distance a train is coming at you okay and the train is blowing its horn or a giving a whistle then you see that as the train moves towards you the frequency that changes okay so that's the Doppler effect so the source is moving here so or if the source is at rest and the observer is moving you know the observed frequency of sound is also different so we want to look at whether Doppler effect is there in light or not well that's the thing for the next lecture thank you very much.