 So, today we are going to talk of the consequences of special relativity. I am sure this portion is not new to you, but we will talk of the Doppler effect in light. The prerequisite for that is of course, you know a bit of time dilation things that we have covered earlier. I am sure you have heard of Doppler effect in sound. Let us see what it is there in light when we talk of electromagnetic waves, example light. And after that we will check or rather we will see whether the mass of a body actually whether it varies with velocity we may have heard of such things, but let us see where it comes from. So, let us first of all consider you know we consider a light source and then that is emitting light of a certain frequency. Let us just let us call that frequency new 0. And you have an observer here we show it by O here on to us right of your screen. I mentioned that of Doppler effect in sound. I am sure where instead of this light source you will be having someone some source which is emitting sound, which is making some noise and then emitting certain frequencies. And then it is all fine when they are at both of then both the observer and the source is at rest to teach other, but what happens if all of a sudden one of them starts moving. So, let us say the one of the source start moving and then the observer starts moving. Well, if you have the observer starts moving towards these the source of the sound do not you feel that you know the frequency of the sound that you are hearing has increased. It is more like a car coming towards you you know and it blows the car blows its horn. Then the more it comes towards you you find the pitch of the sound increasing. Well, so what is it with light? Do we observe certain the same things or is it a little bit different? Let us check that out. Of course, need to mention the very beginning. So, for propagation of light here you do not need a material medium because it can propagate to vacuum, but for sound of course you need a material medium. So, that is what we do we consider light source which is emitting light of a certain frequency that is new 0 and then we have an observer. Now, what we can also do you can also consider this a light source you know as a clock that is ticking like this tick tick tick and then at at each tick each one of this ticks and it is emitting a light wave. So, it is so what you are considering you are considering a light source as a clock in a sense that is ticking new 0 times per second and so on each tick it is emitting a wave. So, it is now why do we put all these clocks and things here then because if you put this clocks you know time dilation you really you can make the correlation and then we can use the all the principles of relativity that I have learnt earlier. And now let us see what happens if you have the source here you have the light source which is emitting a certain frequency and observer starts moving. Now, we can have three situations here either the observer is moving transverse your perpendicular to the direction of these light waves or it is moving towards a light source or it is moving away. So, let us let us consider each of them one by one ok. So, case number one. So, in this what we have is that the observer that is we denote by O here and it is moving uniformly with a certain velocity V transverse to the direction of the light waves. Now, if you are you know in the frame of the light source itself where the things are traced and what is the what is the proper time in the sense between the ticks you know you consider the light source to be a clock which is sticking and each of the stick it is emitting a it is a it is emitting a light wave. So, what is the proper time here it is called a T0 then how is it related with the frequency? Well you know you know frequency is in hertz and time it is in second. So, it is one by new 0 ok. So, that is the proper time interval between the ticks in the light source frame ok. Now, what about how would the observer then find this time to be ok. Now, in the frame of the observer the time elapsed between two ticks would be T. Now, T will not be same as T0 which was the proper time remember, but T will be T0 divided by root over of 1 minus V square by C square ok. It is this very simple time dilation formula that we have used here ok. Now, what would then be what would then be the frequency of the light as observed by the observer. Well you take just the inverse of the time here remember this time is the one measured by the observer in his or her frame ok. So, let us call this frequency as new T it is going to be 1 by T and then you know what 1 by T is you are going to you just see in the previous paragraph. So, you will see that it is root over of 1 minus V square by C square whole divided by T0. Now, T0 is also related with the frequency at which the light is being emitted ok. So, you see that. So, new T that is the frequency as measured by the observer that will be new 0 times root over of 1 minus V square by C square. Now, since V is always less than C here ok, it is going to be VT or the since the velocity of the of the observer is always going to be less than the velocity of light ok. So, what you find as new T that is the frequency as measured by the observer will always be less than the proper frequency ok proper time interval and then you know proper frequency if I can put it that way. So, new T here is less than new 0. So, that is the frequency that is that will be measured by an observer who is moving transversely to the to the source of light ok. So, that leaves us with two other cases one in which the observer is moving towards the towards the light source and in the other when the observer is moving away from the light source ok. So, let us consider the other one first as a namely the one in which the observer is moving away from the light source ok. So, here to you know what is the proper time between ticks it is going to be 1 by T 0 which is like 1 by new 0 that is when new 0 is the frequency at which the light is being emitted at the source ok. And remember here the observer is moving at the velocity V away from the light source. So, in the frame of the observer if T is the time that you know small t is the time that the observer will measure between each of the ticks. So, as to say then for in time T remember the observer is also moving with a certain uniform velocity V. So, the observer has already travelled V T that is that is the unit of distance here away from the source between the ticks ok. So, this implies that the time interval between two successive ticks as measured by the observer will be what will be V T by how much will it increase will increase by V T by C where T is the time as measured by you know by the observer here right. And V T by C and C is the velocity of light and that has the same you know that. So, the value of C is going to be the same in both frames ok. Now, what will be the total time between the arrival of successive waves or the successive light waves in the frame of the observer? It will be first the time between two ticks as measured by you know by the observer plus the amount of time you know the observer has moved away in a sense by which time the observer is already moved by a certain distance between these two ticks ok. So, the time lapse between the arrival of successive waves will be capital T which is small t plus V T by C now you simplify this as T plus and then you take a bracket of 1 plus V by C ok. Now, how is T related with the proper time small t that is small t is related with the proper time T 0 by small t being equal to T 0 by root over of 1 minus V square by C square and then you have the bracket 1 plus V by C. So, that simply simplifies to the one given at the bottom of your screen that is T 0 into root over of so, 1 plus V by C divided by 1 minus V by C. Now, how would you find the you know the frequency here? It is that all you need to take is the inverse of the time interval in this case. You see here I have put it as V nu minus. So, that is 1 by capital T. Now, that you just invert the expression that we see in the top of your screen and then you relate T 0 with the frequency that is the frequency with which the light is being emitted by the light source and then you immediately find what is a nu minus and then you can relate nu minus to nu 0 and then you see that nu minus is actually less than nu 0 here. So, in every case you see that the change in frequency so far we have seen that it is depending on the velocity of the observer, the velocity with which the observer is moving away from the light source here. So, what about the last case? So, in this case let us say the observer is moving towards the light source. So, by the same logic which we had applied earlier, the observer travels a distance which is V times small t towards the light source between successive ticks of the clock that we have considered here. So, the total time interval between the arrival of successive waves here, light waves here would be that is the capital T would be small t and then remember it is minus V t by c. Why minus? Because the observer is moving towards the light source and then we have kept the velocity of light is the same in both frames. I mean whether you are in a moving frame, uniformly moving frame or in a rest frame. So, that is why it is the same c. So, it is natural now. So, you do the simplification. You convert the small t into t0, t0 being the proper time. And here what we will find is that when you take the observed frequency, when you find the observed frequency, it is going to be, I have put V nu plus here. That is nu plus is 1 by capital T and we are going to see that nu plus here is more than nu 0. It is going to be more than the frequency of the emitted light. So, we have studied three cases in which the first case the observer was moving perpendicular to the light source. Then the one was moving away from the light source and then towards the light source. However, I should also mention here that in all cases you have seen that in all these cases the shift in the frequency as observed by the observer depends on the velocity of the observer here. So, you do see it in all this formula here. Another important point that I should mention here is that in each of the formulae that you have seen so far, I have always considered the light source to be at rest. Why? Because had I not for example, the light source was moving and then the observer was at rest. That would have been a similar situation because it is the relative velocity which is important here unlike the case of Doppler effect in sound. So, it is not only in the Doppler effect in sound, it is not the relative velocity. You have to take into consideration the velocity of the speed of the source and speed of the observer here. But here what we have it is the relative velocity of between the observer and the source which is going to be important here. So, if the source is moving or the observer is moving it does not matter as long because the velocity that we are considering here is actually the velocity V that you see in all the formulae will always be the relative velocity. So, having done this, let us spend a couple of minutes on where possible applications of this derivation would be. Well, as I told you, you see in the shift in frequency is related with the relative velocity with which the object is moving. So, that immediately told people that maybe it could be used to measure the speeds of distance Taylor objects like stars. So, what do people do? What do astronomers do in that case? What they do is that they take a photograph of the spectra of elements which is in the star. So, from the light which is emitted from the star the light spectra is analyzed. So, and where the frequencies are people find out and then they compare it with the known spectra of elements present in the star. So, these known spectra of elements you can find the spectra of elements by some experiment at the laboratory here on earth and then they compare that spectra with the spectra that was obtained from this star. Now, if the star is moving away or there is a relative velocity, there is a relative motion between earth and the star there and then the star we are going to have a shift in spectral frequency. Now, from the shift in spectral frequency you can then find the speed of the star. So, it is got lots of applications in astronomy of course, then many other some other discoveries were also based on this. Another application actually a very interesting application of this would be to measure the temperature of hot plasma in nuclear fusion experiments or the temperature of very hot gases. Now, what they do is that well the principle is almost similar you know to what I had what I have outlined earlier it is that they take the spectra of you know when things are moving at such high speed. So, they are emitting radiation and this radiation is analyzed and then if it is a known gas. So, from the known spectra you can find out what is the shift in the spectral frequency and then you can relate that with the speed of the molecule of the gas which is there inside your experiment and then if you relate if you remember you connect theory of gases. So, from this velocity or you know the mean velocity or the root mean square velocity you can always relate that to the temperature that can always be done from the kinetic theory. So, that will give an estimate of the temperature involved here. So, we have studied Doppler effect in light as against you know that now as against Doppler effect in sound and then we have also talked a bit about a few applications. Fine. So, now we are going to shift gear a little bit and do something a little bit different and go to another consequence of special relativity and study if the mass of a body actually varies with the velocity with which it is moving. When I know that this is one of the things that many of us are rather aware of when we study physics or any of the elementary courses of science we take mass increasing with velocity and equal to m c square but all of it in due time. To do this problem of where the mass of a body varies with velocity let us start with an example and then we will come to a conclusion. So, concentrate on the thing written down as s prime frame. So, what we have here is that we are considering the inelastic head on collision of two identical particles a and b in s prime frame let us say. So, what is this s prime frame? Well it is a frame which is let us say it is moving with a certain velocity certain uniform velocity v with respect to another frame s let us say. Now in it we have the inelastic collision of two bodies here named as a and b. So, they are of identical masses m prime both are of m prime and they are moving with the same speed you know opposite to each other and then they are heading towards a head on collision that means speed is u prime here. We have all put primes here because you know to keep in to keep you know in to make since we are talking of s prime frame. So, we are dealing with prime coordinates here let us put it that way and then all these velocities u prime you know they are moving parallel to the x axis. Fine so, they are identical masses also. Now how is the same event observed in s frame? Well you have different you have you know in the s frame you are going to see the same thing as some you know the velocities may not be the same. The velocities although they are moving the the speeds are all the same, but if you take the modulus of the speeds in s prime frame they are the same the directions may be different, but the speeds are the same. But then the velocities of the speeds are not going to be the same in s frame for a simple reason it is that you need to invoke Einstein's velocity addition formula to find what is the velocity in in s frame. And you go to see immediately that it is not going to be the same it is just written in you know in for your convenience at towards the end of the slide. For example, if you just consider u 1 so, of the mass of the velocity of mass 1 mass a rather you are going to see in terms of the s prime quantities and then the velocity v it is u prime plus v divided by 1 plus u prime v by c square. However, if it was minus u since was moving in other direction u 2 that is the other you know the other quantity that is minus u prime plus v into 1 minus u prime v by c square so, all going to be the same. So, you might be a little bit confused why you have written different masses for identical masses. So, in the s prime frame you had identical masses both are m prime. But why is it that they are different in in the s frame? Well at this stage I do not know why they are different rather I just take them to be different. So, just to keep myself in conjunction with the idea that I have different velocities I do not know what is going to the masses. So, if it is if my if my from my relativity from whatever I do it turns out from all this conservation laws I am going to invoke on if it turns out that these masses are the same they will turn out to be same m 1 will be equal to m 2. But right now I am going to take m 1 and m 2 different. Let us just see what happens. So, that is what we have we have the s prime frame and then in the s frame we are going to see what it is going to be of the same even the same collision how is going to how how one is going to see in s frame. And remember s prime frame is moving with a certain uniform velocity v with respect to the s frame. So, on top of that of course, on the back of my minds we we have the conservation of linear momentum in our in the collisional process. Well this is rather something very sacrosanct. I mean we are not going to violate all these conservation principles whether it is non-relativity or relativity or classical mechanics. I mean the conservation laws are conservation laws you are supposed to you are supposed to you are supposed to maintain them. And then on top of that you also have the total mass to be conserved that is the total variable mass you know the total mass. For example, if it is consider s prime frame you have m prime a and then the other mass b also of m prime. So, after collision the mass is going to be m prime plus m prime that is 2 m prime that is what I mean the total mass means conserved in the collision process in each of these frames. So, what does it mean here? Well from the linear momentum conservation what we have in this collisional process if you look at the s prime frame it is very simple it is going to be 0 is not it I mean it is they are moving that your identical masses the speeds are identical and opposite to each other moving parallel to the x axis x prime axis. So, u prime m prime u prime plus m prime minus of u prime that is going to be 0. So, momentum so even if the if so that tells you that if you are going to have an inelastic collision in the end. So, you know the mass is going to be whatever it is and the velocity or the it is it has to be at rest in the s prime frame after the collision. Well something of that later on, but what about this the total final mass of the combination is final mass of the combination m prime plus m prime it is going to be 2 m prime there is nothing rocket science in that, but the only thing is that it is going to be maintained after collision that is what I want to say. So, that is what we have the thing. So, just one look one more slide to to recapitulate on this collisional process before the collision. So, we have the s prime frame where we have identical masses a and b they have similar masses they have the same mass m prime and they are moving opposite to each other parallel to the x prime axis and then one is observing the same thing from the s frame and then the s prime frame is moving with a certain velocity p with respect to the s frame. And from the s frame we are going to see the same collisional process, but then the velocities are not going to be different and then we have taken different masses here we do not know what we do not know whether the masses will be same or different, but let us see if from all these conservation laws what turns out to be what whether we have indeed different masses whether m 1 is separate is different from m prime. So, what happens after collision remember the collision is perfectly in elastic. So, what it means is a combined particle the combined particles will be at rest and they will stick together in the s prime frame. So, that is what they are they will stick together in the s prime frame the total mass is 2 m prime and they are at rest. Now, what will be the velocity with which these things will be seen to move from s frame? So, is it v we are going to find that out. But remember here what will be the total mass of this a b system here at in s frame it is going to be m 1 plus m 2 because total mass is conserved in this collisional process. So, what about the velocity as we said. So, remember after collision well still after collision we are we are still considering s prime frame to be uniformly moving with a certain speed v with respect to the s frame. Now, in s prime frame a and b is at rest right is at rest. So, what does it tell you and then the s prime frame is moving with velocity v with respect to s. So, what is the velocity with which a and b will be seen to be moving from s frame well it is v that is what we have. So, in s frame we have these two particles they are sticking together since it is sticking together in s prime frame it is sticking together it is going to have mass m 1 plus m 2 and it is going to move with a certain velocity v which is same as the relative velocity between s and the s prime frame. So, just to recapitulate the entire collision process what you see in the top of this slide is the situation before the collision both in s prime frame and the s frame just have a nice look on the situation what is happening in the s prime frame and the situation that we have in the s frame that is the situation before collision and then after collision it is perfectly inelastic collision. So, it is sticking together in s prime frame and total mass is 2 m prime and it is at rest and then from s frame we are going to have the total mass to be m 1 plus m 2 the total mass is conserved here and it is moving with velocity v. So, next what we are going to do is that we are going to invoke the conservation of momentum in the s frame and remember if you invoke the conservation of momentum in the s prime frame it is going to be 0 is not it I mean the total momentum total linear momentum in the s prime frame is 0 and what is it if you invoke that same conservation of momentum in the s frame what is it going to be it is going to be m 1 u 1 plus m 2 u 2 that is the total momentum in what in the s frame before the collision and then after the collision it is going to be m 1 plus m 2 whole times v that is the velocity between this combination if a plus b or this m 1 plus m 2 is moving. Now, we rearrange a term do this little simplifications immediately get what m 1 plus m 2 is m 1 plus m 2 is nothing but v minus u 2 divided by u 1 minus v well that is very simple you just take all the m 1s you know all the terms involving m 1s on one side all the terms involving m 2 on one side and then do simplifications. And then what you do is that you substitute for u 1 and u 2 the expressions from Einstein's velocity addition formula. So, you know that very well. So, u 1 is 2 prime plus v divided by 1 plus u prime v by c square and what is u 2 u 2 is minus u prime plus v divided by 1 minus u prime v by c square. So, you are going to substitute these two velocities u 1 and u 2 in the expression given in the middle of your screen then given by this m 1 by m 2. Well, I am doing a little bit of the algebra a bit rather little bit of the algebra which only but if you miss this slide as a next we are not going to miss most of the physics. I am just doing it so that you will have the expression you can derive all these things by yourselves. So, what I have done is that now what will be m 1 plus m 2 in terms of all the quantities in the prime frame. So, it is 1 plus u prime v by c square divided by 1 minus u prime v by c square again what we do is that. So, we have derived up to the expression at the top of this slide again what we see is that a little more simplification actually you can derive this if you invoke again the Einstein's velocity addition formula. 1 minus u 1 square divided by c square and what is 1 minus u 2 square divided by c square. So, you put in what is u 1 and u 2 in terms of u prime v's and all these things you see rather very interesting expressions you can you can you can actually check that out. It is 1 minus u prime v by c square whole square divided by 1 plus u prime v by c square whole square. Now, from the top of the slide you can see that this is nothing but m 2 by m 1 whole square. Now, check what we have check the 1 on the left hand side. So, it is 1 minus u 1 square by c square divided by 1 minus u 2 square by c square and then on the extreme right hand side you have this m 2 by m 1 whole of square. This is very interesting these are the two expressions of the left hand side and the right hand side disregard the one in the middle. What do you see? You see the masses you see the velocities. What do you again see? What you see is m 1 into 1 minus u 1 square by c square and you take a square root of that is equal to m 2 into 1 minus u 2 square by c square take a square root of that and that is it varies. That is not changing. So, what you see is that what you have here? So, m 1 and m 2 are the masses of identical particles. When the velocities are u 1 and u 2 respectively I mean when I say m 1 and m 2 are masses of identical particles and you know does not carry much of a meaning. I mean then you could say that identical particles how come they have different masses. But you see that they are moving at different speeds. So, this combination some rather interesting combinations of m 1 and u 1 of these speeds with which they are moving and then use that combination given by m 1 times 1 minus u 1 square by c square that square root of that that is a thing that is going to be invariant. Change the velocity the expression for the mass changes that is becoming invariant. So, what is invariant is this combination in whatever you know whatever mass and then it is corresponding velocity that is possible here. So, what we have? So, in a frame if v is the velocity of the particle the v is the velocity of the particle and m is its mass then the quantity that you will have m into 1 minus v square by c square root over of that that is going to be invariant. Now, this is very interesting why now what is this well rather what should be the invariant quantity let us check that out. Now, if you take the velocity with which the mass is moving to be 0. So, it is basically the mass is at rest. So, you consider you consider a frame in which the particle is at rest and then you measure the mass there. Now, if the mass is m 0 there. So, you plug in this quantities into the formula. So, m into 1 minus v square by c square. So, the mass is m 0 and the velocity is 0. See the 1 in the bracket 1 minus 0 square by c square and the root of that. So, that is going to be 1 no problem and then on the right hand side you have m 0. So, that is going to be the invariant quantity and m 0 is a mass of the same particle in a frame in which it is at rest. So, that is the formula we are looking for. See that if you know that the invariant quantity is actually m 0 here you can now relate the mass of a body which is moving with a certain velocity v as m is equal to m 0 which is the mass of the body in a frame in which is at rest and then divided by root over of 1 minus v square by c square. So, what we have is the mathematical formula and then this is very interesting of what the mass of the body will be if it is moving with a certain velocity v. Now, let us see how this is related. So, let us have a graphical illustration of this variation of mass and velocity. It is if you look at the graph on your screens what I have plotted here on the y axis is the mass and on the x axis I have v by c. It is basically it is the velocity in the units of velocity of light. So, it is someone it is starting from 0 to the speed of light. So, when v by c is 1 it is the speed of light. So, when v by c is 0 then you know that the body is at rest. Now, if we consider a mass of which is at rest which is m 0 to be of let us say 10 kg. Now, that is the green line. So, that is the green line and so, whether if you increase the velocity or not. So, that is m 0 it is the it is the rest mass. So, that is 10 kg, but as you increase the velocity see the red see the red curve see that at small speeds when the velocity of when the velocity of the body is not too high or it is too high not too high compared to the speed of light you see that it is it is increasing it is increasing and then all and then when it when it reaches towards the speed of light you see all of a sudden it increases very fast I mean it is almost explodes. So, you do see that the mass of the body is increasing in a certain way when when when the velocity is increased. Now, I also wish to remind you of one more thing it is that when you are at very small speeds when on basically when you are non-relativistic speeds when v is much much less than c you can immediately figure out look at the formula. For example, let us first look at the formula. So, m is equal to m 0 divided by root over of 1 minus v square by c square. So, when is much much when v is much much lesser than c v by c almost tends to 0. So, the denominator in this fraction is going to be 1. So, m will roughly be equal to m 0 when the velocity is much much less than the velocity of light and that is exactly what you see in the graph. So, when velocity is much much less than the velocity of light you are in the non-relativistic limit in the classical limit the classical world that we all live in you see that the red line actually converges with the green line. So, it actually fits with our experience. So, now let us see so so let us let us see what have another example of what the consequences of this of this formula would be. Let us just consider the rest mass of an electron the rest mass of an electron if we look up the tables it is something like 9.11 into 10 per minus 31 kilograms. Now, suppose this is moving with it is a pretty high velocity it is 80 percent the speed of light. So, will its mass be equal to you know how much will it mass what will it mass be will it be very near to 9.11 and 10 per minus 31 kilograms let us see. So, that is the rest mass and the velocity V here is 80 percent of the speed of light. So, what you see here will be m so that is the measured mass will be equal to m 0 divided by root over of 1 minus V square by C square. Now, m 0 you know what that is 9.11 into 10 per minus 31 kilograms and then 1 minus 0.8 square C square divided by C square. Now, this turns out to be something like 15 into 10 per minus 31 kilograms. So, you do see an increase in mass when you when the velocity is increasing and that too when the velocity is much much well it is actually 80 percent the speed of light it is quite a huge amount of which is quite a large velocity. So, by hope today I have you know you be able to cover some part of the consequences of special relativity we talked of Doppler effect in light and then the interesting phenomenon of the mass of a body varying with its velocity. In the next section what we will talk of is the relation between mass and energy and I will heavily draw upon what we have done so far about the consequences of relativity especially about the variation of mass with its velocity. Well, thank you very much.