Hello, in our last lecture, we discussed the concept of photon. We said that special theory of relativity gives us a possibility that a particular particle may have zero rest mass and may still carry energy and momentum. In classical mechanics, we can never image in a particle with zero rest mass, but special theory of relativity gives us that possibility. And we discussed saying that the light can also be imagined as consisting of particles which have zero rest mass and they are called photon. Of course, we have a condition that if the particle has a zero rest mass, it must travel with speed of light and we know that light travels with speed of light. So, this is consistent in saying that light consists of particles which has photons. So, this is what we have said. We discussed how light can be thought of consisting of particles known as photons. We also discussed Compton fact in our last lecture, which was one of the experimental proof in which we treated photon like just any particle and assumed that this gets collided or get scattered with an electron and treated just this as like we would have in classical mechanics discussed any other collision process or scattering process except for the fact that we utilize relativistic expressions. Then we found that the frequency or the energy of the photon after scattering would get changed and we could evaluate that depending upon the angle of scattering what will be the changed wavelength or change frequency or change energy. Then we described an experimental way of determining this change and we found that the two are consistent. In that sense, Compton effect is a very interesting experiment because it provides us a proof, so to say proof that light can be really thought of consisting of particles which have zero rest mass, which carry momentum and energy like any other particle. Then we worked out an example where we saw that energy and momentum of photon can be transformed using standard energy momentum transformation. See, as far as relativity is concerned, there is nothing special about photon except for the fact that it has a zero rest mass. There can be particles with different masses, different rest masses. Similarly, photon is also a particle with zero rest mass. Therefore, it does not require any other special treatment other than the fact that m naught is equal to 0. So, all the transformation expressions that we have derived earlier should also be valid for the case of photon. And we dealt with an example in our last lecture where we really saw that we can apply exactly the same type of energy momentum transformation that we can apply to any other particle. We can also apply to the case of photons. Now, let us go ahead and discuss what we call as a Doppler effect. If we change a frame of reference, we know that the speed of light will not change. That is what we have always said. This is one of the postulates of relativity, but we have just now said that energy and momentum will get transformed. It means they would change if the frame of reference changes. And what energy and momentum is related to in the case of photon is by this expression e is equal to h nu, energy is equal to h nu and momentum p is equal to h nu by c. It means change of energy, change of momentum implies a change of frequency. It means if I change my frame of reference, then this is equivalent of saying that the frequency of the photon has changed, though its speed has not changed remember. So, its frequency has changed and this change of frequency once we go from one frame to another frame is called Doppler effect. So, that is what we have written. If the energy and momentum of the photon changes upon change of frame, this would imply a change in frequency in the other frame. This is called Doppler effect. Doppler effect is very well known in sound. In fact, the first example that we have always been giving of Doppler effect is in sound, which many times we see in our daily life that a train is coming towards us. We find as if the frequency of this, you know its whistling, the frequency of the whistle from the train is appears to be of high pitch, higher frequency. Once the train recedes, we find that this frequency is of lower frequency. This is a well known Doppler effect and very well explained understood by the wave theory in the classical mechanics case. So, this is similar to that, but there are certain different ways of treating it because of the fact that I am going to describe next because of the relativistic effect, the light has to be treated somewhat differently from the sound Doppler effect. Let us just recapitulate the sound Doppler effect, then we come to the actual expression of the light Doppler. In sound Doppler effect, we get two different expressions. Let us take a simple case. One case in which observer is stationary and the source is moving. Another case in which source is stationary and observer is moving. These two cases in the case of sound are treated differently because of the main reason that sound requires a medium to travel. So, whenever I am talking of the speed of let us say source which is emitting sound or an observer which is hearing that sound, the motion or the speed or velocity of these source and the observer are always defined with respect to the medium. And these two expressions, these two cases when observer is moving and the source is at rest and in which source is moving and observer is at rest corresponds to different cases. We will explain this in a very simple fashion. Suppose the observer, I am sorry the source is at rest and this is the entire air medium in which it is emitting sound waves, then all these sound waves will move out in the spherical wave front from this particular source. Now, these wave fronts, these are sort of circles and I have not drawn them very well. So, these are sort of circles. Now, these spherical wave fronts move in the air medium because sound requires this medium to travel. Sound is essentially a mechanical wave in which this medium which requires this medium. In fact, these vibrations are carried forward in this particular air. If we do not have this air or any other medium, we will not hear the sound. So, this particular medium which I am assuming to be air, these waves are all generated. These wave fronts are created in this particular medium. If a person is walking away or walking towards this or walking away from this, then the rate at which it is sort of listens at which it comes across these wave fronts is going to be different from the original cases. Now, if we have a different situation when this particular emitter of sound or the source of sound itself is travelling, but the person is stationary, we will find that these particular wave fronts, they themselves are created in different fashion because initially when the source was here, it created a wave front which is something like this. And when it moves towards this particular person, next time the wave front will be centered according to this person. Here, remember with respect to the medium. This person has travelled from here to here. So, now the wave front would be circular with this particular circle, this particular point in the center. Sorry, my figure is not very good. Very good. But essentially, it means the way the disturbance is created inside the medium that itself is different. So, the two situations are different and the expression of change of frequency that we get corresponding to these two cases that also turn out to be different. So, this is what I have written. The expression of change of frequency as far as the Doppler effect of sound is concerned depends on whether the source or the detector which is moving. The difference is caused because sound needs a medium to travel. The speed of source and or detector are defined with respect to the medium. The expression of change sound frequency when source is fixed and observer is going away with the speed V is given by this means if we have source which is fixed and this observer which is going away from here like say like this or observer is going like this. If nu naught is the frequency which is being emitted by this particular source nu naught is the frequency which is emitted by this particular source. The frequency which should be heard by these two persons persons which are moving in opposite direction their distance is increasing will be given by this expression nu ds is the change frequency which I am calling as Doppler shifted frequency is equal to mu naught which is the original frequency which is being emitted by the source multiplied by 1 minus V upon C where V is the speed of the observer the speed with which it is going away remember this speed is relative to the medium. So, that is what I have said the expression of change sound frequency when source is fixed and observer is going away with the speed V is given by this is the nu frequency this is the frequency which will be heard by the person. On the other hand if the source itself is moving which I have said is fundamentally different from the earlier case then I will get slightly different expression and this expression will be given by nu ds which is Doppler shifted frequency is equal to nu naught which is the original frequency divided by multiplied by 1 divided by 1 plus V by C where V is the speed with which the source is going away from the observer. So, this is I have written the expression of change sound frequency when the observer is fixed and the source is going away with the speed V is given by this. Now, if you are you can probably realize that the case of light has to be different because as we have said of course, in our olden days we have been talking about ether medium that light requires an ether ether medium to travel, but special theory of natively discarded that concern we do not require a medium for the light to travel. Therefore, we cannot create a situation when the light source is coming there is no medium in which these wave fronts are created. Therefore, whether the observer is moving towards source or source is moving towards observer these two things in relativity have to be treated exactly identically they cannot be any difference between these two because these two speech all that we can talk is the relative speech between the two there is no medium in which this light is travelling unlike the sound wave. Therefore, I must get the same expression in both these cases when we are talking of Doppler shift of light. This is what I have written for light the two situations are fundamentally similar as it does not require a medium to travel. Hence we expect a single expression to represent both the situations the two situations cannot be different. I hope you can understand the difference between a light and a sound requiring a medium to travel. Therefore, I can define the velocity of an observer and a source relative to that medium while light does not require a medium to travel. So, if I talk of velocity I can only talk of relative velocity within observer and the source I cannot talk with respect to anything else because there is no ether. So, let us first consider a case of what we call as a longitudinal Doppler effect. This is the picture which gives what is called a longitudinal Doppler effect. Let us first spend little bit of time in understanding this figure. So, this figure is essentially simple. We have just to fix our idea we can imagine that this source this particular person S frame is a ground frame of reference and assuming that ground is an inertial frame of reference just to fix our idea it is always simple to imagine things and there is a train or a carriage which is moving towards right with a speed or with a velocity V. This particular light or this particular sorry this particular carriage or this particular train emits light in backward direction. So, remember the figure the way we have drawn in case the light has to reach this particular observer S which is sitting on the ground it has to be emitted only in this way if it is emitted this way this will not be seen by the observer S here. So, I am imagining in this particular case that the motion is collinear the observer S is sitting on the ground it is another observer S which is standing on the train which through some hole or somewhere shines a light in this particular direction. So, that this particular light is seen by this particular observer. Now, when I say source frequency essentially it means frequency which is seen in the frame of reference of the source here everything is contained in this particular carriage train carriage which I am calling as S frame of reference. So, I can say that this frequency new prime is the frequency of light as being measured by observer S. So, its frequency which is measured in S frame of reference or in other words this S new prime is the energy of the photon as measured in S frame of reference. Now, my question is what will be the frequency which will be seen of the light by this particular observer S and this particular change frequency will be what we will call as Doppler shifted frequency. So, all I have done the same problem of Doppler effect I have changed the language into the relativity language or the language which we have been using of different frames. Now, if I have to find out what is the frequency of this particular light in this particular frame S essentially I want to find out what will be the energy of this particular photon in S frame of reference all I need is a energy momentum transformation. Once I do the transformation of energy I will know what is the energy of the photon as measured in S frame of reference. Remember, we have said photon is like any other particle. So, when we can apply momentum conservation when I go from S prime to S or S to S prime for any other particle I can equally well apply also for the photon. So, my problem is very straight forward in fact Doppler effect in light can be treated extremely simple simply all I have to do is to apply an energy transformation. So, that is what I am doing in the next transparency. So, I have applied energy transformation and if you remember the energy transformation I am applying a inverse transition this energy E in S frame will be given by gamma multiplied by E prime where energy E prime is the energy in S prime frame of reference plus V which is the relative velocity between the frames and P X prime is the momentum of the particle in S frame of reference. Now, we have just now said that the energy of the photon as seen in S frame of reference is H nu prime. Therefore, I can write E is equal to H nu prime. Only thing I would like you to pay attention to the fact that this particular photon has to be emitted in minus x direction. So, it reaches the observer S. Just look at the earlier picture this photon has to be this is the direction of the relative motion. So, this is my plus x direction this photon is being emitted in minus x direction. So, as to reach S therefore, this momentum must have a negative sign it must be minus x direction. Therefore, I am writing P X prime as minus H nu prime by C this what I have done I have written P X prime is equal to minus H nu prime by C because if this momentum was this particular photon was travelling in a plus x direction then it will not be seen by that. Of course, we can create various type of situation the final expression will always turn out to be the same. This situation which I have created is the one in which this carry is moving forward and the photon is moving emitted backwards. So, that it can be seen by an observer S. So, I write P X prime is equal to minus H nu prime by C and the energy as seen by an observer S will be termed as H nu. So, E is equal to H nu. So, in this particular expression I substitute E is equal to H nu and E prime is equal to H nu prime. So, this was my original expression E is equal to gamma E prime plus V P X prime for this E I will write H nu which is the change frequency this remains as gamma this E prime I will write as H nu prime this V is in plus x direction, but this P X prime is in minus x direction. So, I will put a negative sign P prime P X prime will be given by H nu prime by C multiplied by V. Therefore, if I cancel H I will get nu is equal to gamma nu prime minus nu prime V by C. So, this is what I am writing in the next transparency. So, I have just substituted H nu is equal to H nu prime minus V H nu prime by C and this gamma have expanded written as under root 1 minus V square by C square. Canceling these H we get an expression nu is equal to nu prime because I have taken this out. So, this will become 1 minus V by C divided by under root 1 minus V square by C square. So, this is what I get the expression remember nu prime was the frequency in S prime frame of reference nu is the frequency in S frame of reference. Now, this under root 1 minus V square by C square I can write as under root 1 minus V by C under root 1 minus 1 plus V by C 1 plus V by C. We just factorize A square minus V square is going to A minus B into A plus B. Now, in the numerator we had 1 minus V upon C. So, this will cancel with this root of that will cancel with root of 1 minus V by C giving me this particular expression which I have written here nu prime nu is equal to nu prime under root 1 minus V by C 1 under root 1 plus V by C. So, remember this root of this has been cancel with under root 1 minus V by C under root 1 plus V by C still remains here and on the numerator there is an under root sign remains here. So, the nu expression that I have got is nu is equal to nu prime under root 1 minus V by C plus divided by 1 plus V by C. Now, what I would like to just mention that in Doppler effect we try to use a slightly different type of notation. So, let us not get confused with that particular thing and see remember in relativity language we call nu prime as the frequency or h nu prime as the energy measured in s frame of reference and nu as the frequency or h nu as the energy measured in s frame of reference. But in the language of Doppler effect the frequency which was emitted by s prime we call that as original frequency which we call as a nu naught which is the frequency of the source and the frequency which we have seen in any other frame we call that frequency as the Doppler shifted frequency. So, I am just changing this particular thing I am changing nu to nu ds and I am changing nu prime to nu naught. So, if I do that this is what I call as a Doppler effect notation I get the same expression just changing this particular thing nu ds is equal to nu nu naught multiplied by 1 minus V upon C root 1 minus V square by C square same expression which I wrote earlier which is equal to nu naught under root 1 minus V by C divided by 1 plus V by C. This is the original frequency this is the Doppler shifted frequency same thing has been re-dominacled depending upon the way we want to talk. In Doppler effect we talk of the original frequency and the shifted frequency while in relativity we talk of s frequency in s frame and frequency in s prime of this particular expression can be simplified little bit if we assume that the relative velocity between the frames is comparatively small in comparison to C. As we will be giving some examples today we find this particular Doppler effect in light to be quite useful in many of the experimental techniques that we use. So, and in many of those cases the relative velocity between the frames is not that large. So, we just approximate this particular expression under the condition that V upon C is very small in comparison to 1. So, that is what I try to show in the next transparency. And remember my expression was 1 minus V by C under root this was in the numerator. So, this can be written as 1 minus V by C to the power of half. In denominator we had an expression of 1 plus V by C this I bring to numerator this expression or let me put this 1 divided by this can be written as 1 plus V by C to the power minus half because this was this was in denominator this already under root. So, once I take to numerator the power becomes negative. So, this I can write as 1 plus V by C to the power minus half. So, when I multiply these two it means I multiply this by this expression I will get 1 minus V by C to the power half multiplied by 1 plus V by C into the power minus half this minus sign has come because this was in denominator. Now, because we are assuming that V by C is very small V by C is very small in comparison to 1 I can expand this into a binomial expansion and neglect higher order terms retain only the first two terms and we will get first to first approximation that in the limit of low relative velocity between the frames what will be the change frequency. So, this is what I have written in this particular transparency. So, V nu ds is written as nu naught is equal to 1 minus V by C to the power half multiplied by 1 plus V by C to the power minus half which is approximately by expanding this. So, this becomes 1 minus let us say n times x or n times V by C. So, n is half which is the power. So, this becomes 1 minus V by 2 C. Similarly, here there is a plus and the power is minus half. So, this becomes 1 minus V by 2 C. So, this minus half gets multiplied here for the first term 1 plus nx. So, 1 plus if you have to expand 1 plus x to the power n this becomes 1 plus nx in the first term. So, this I can now multiply neglects second order terms. So, 1 multiplied by 1 will give me just 1 which is this one when this minus V by 1 T C gets 2 C gets multiplied by this one I will get minus V upon 2 C. Similarly, this minus V upon 2 C where it gets multiplied by this one I will get another minus V by 2 C when I add these two I will get 1 minus V by C. One can just simply work out this is a very simple way of just opening up multiplying these brackets. Of course, we will have a term of V square upon 4 C square which I will neglect because this is a second order term in V upon C. So, I find that Doppler effect effected frequency is given by approximately nu naught multiplied by 1 minus V by C which happens to be the same expression which we have derived I mean not derived, but for the case of sound when we have used the expression for the case when your source was fixed and the man was. Now, there is another type of Doppler effect which does not happen in sound which is called transverse Doppler effect which is purely a relativistic phenomenon. See, remember in longitudinal Doppler effect we assume that source is here and the train is moving just in the same line. Now, look at the situation which is given in this particular figure. Let us assume that this particular person S is standing here and this train this is a train track in front of him and this train is going like that and this particular train is emitting sound in a direction perpendicular. Remember when I say perpendicular, let us be little careful. When I say perpendicular it is such that as for this particular observer S is concerned to him it appears that this particular is not really sound this is light I am sorry that this light is travelling towards him. So, essentially it is a situation which is something like this that this particular train is moving like this. Let us just represent it by one single point here and this observer S is here and this observer S is here. According to S when this particular observer was just perpendicular to this particular direction of motion light is emitted towards him as seen by observer S. Remember according to S, this particular light will not really move perpendicular to his motion but this particular observer S receives the light towards him in a direction perpendicular to the direction of the motion of the train. Remember observer S is observing the motion of the train in this particular direction and the same observer S finds that light is travelling in a direction perpendicular to the motion or perpendicular to the velocity direction and it reaches him. So, this is somewhat the situation which it sees. Now we can find out using again energy momentum transformation that if this is the frequency or this h nu prime is the energy of the photon which is emitted by this S observer what will be the frequency of this particular photon as will be seen by the observer S. This is what we call as a transverse Doppler effect. In sound we do not see such type of transverse Doppler effect this purely relative step phenomenon and let us first work it out. We use a direct energy transformation which means E prime is equal to gamma E minus V times Px because this is a direct transformation. Therefore, there is a negative sign there is no positive sign. Remember I have used direct transformation because I know that Px has to be 0 because remember x is the direction of relative velocity and according to an observer in S frame this particular photon is moving purely in y or minus y direction perpendicular to x direction. Therefore, Px is 0. Let me come back to this particular figure which I have drawn here in this particular paper. So, here there is a S which is this motion is in this particular direction. So, this is my x direction. Now, according to this observer S this photon is travelling in this particular direction. Therefore, according to observer S the x component of the momentum of the photon is 0. Now, in S prime if he has to throw a photon towards it I am using the word throw if it has to emit a photon in this particular direction this particular photon because this is moving relative to S would not be exactly in a direction perpendicular to him. So, let us not get confused it is not Px prime which is 0 but it is Px which is 0. So, E prime is equal to gamma E minus V Px and with Px is equal to 0. Of course, using exactly the same way E prime is equal to h nu prime and E is equal to h nu. So, I put for E prime h nu prime and for E is equal to h nu Px is 0. So, I just get nu prime is equal to gamma nu. Again, I change these two by a relativistic my Doppler fact notation. This nu prime was the original frequency nu naught this nu is Doppler shifted frequency because this is being observed by observer S. So, I will call this as nu dS this I will call as nu naught. This is what I get the expression nu dS is equal to nu naught under root 1 minus V square by C square. So, this is as I said a purely relativistic effect called transfers Doppler effect. A few comments if you look at the expression that we have obtained for the case of light for example, for longitudinal Doppler effect. We can very easily see that if the light source is moving towards the observer its frequency will go up if it is moving away its frequency will go down. Remember the expression that we have found out was nu dS is equal to nu naught 1 minus V by C. Let us show you this particular expression approximately. Now clearly if V is positive it means remember we have taken the motion to be in minus x direction. It means if the source is moving away I am sorry if the source is moving away from the observer then in this particular case 1 minus V by C will be smaller than 1 and therefore nu dS will be smaller than nu naught. Therefore, if the train is going away from the observer we will find the Doppler shifted frequency to be smaller. If it is moving towards him then this V has to be change sign this will become 1 plus V by C and in that case nu dS will be larger. So this is in that sense similar to what we generally see in the case of sound. Of course expression is also very similar to the sound. So there is nothing very surprising about it. Only thing which I want to tell you that as far as the transverse Doppler effect is concerned in this particular case the change frequency will always turn out to be smaller. Now let us give some comments for the Doppler effect. In the case of longitudinal Doppler effect what we found that in the limit of V less than C the expression is similar to the sound Doppler effect. So it is quite clear like the normal sound Doppler effect. If the source is moving towards observer the frequency of the light will appear to be increased. Similarly if the source is moving away from the observer the frequency of the sound will be appearing to be less. So this is what we had written as the expression nu dS is approximately equal to nu naught multiplied by 1 minus V by C which is as we have said expression similar to what has been observed in the case of sound for of course a specific case. And therefore we expect essentially the result to be identical that when V is positive nu dS is smaller than nu naught when V is negative it means the source is moving towards the observer then in this particular case this nu naught gets multiplied by a factor larger than 1 and nu dS turns out to be larger than nu naught. So light we see exactly similar effect. But in the case of transverse Doppler effect we always see a reduced frequency the observed frequency is always lower than the source frequency. This is what I have written here. In LDE means longitudinal Doppler effect the observed frequency is more if the source is moving towards observer and lower when the source is moving away. In transverse Doppler effect the observed frequency is lower than the source frequency as we have said it has no classical analog. I would like to just mention here is that the transverse Doppler effect can also be derived or can also be thought to be as a cause of time dilation. Let us see how if this is a carrier let us draw a carrier or let us just draw a point source to be more specific and this point source is moving in this particular direction. Now this particular source is emitting light you can always think we know that light is electromagnetic wave. So when we say in a wave there is a maximum there is a minimum there is a maximum amplitude you know displacement as we call it in the mechanical wave here in terms of electromagnetic wave this will be a point where you have magnetic field which is highest or electric field which is highest or whatever we want to talk about it. There is a maximum and there is a minimum so displacement is maximum and minimum that is what the wave is like. Now difference between two consequent maximum one after another the difference of time difference between these two is what we call as a time period in a classical wave.  There is another wave which is emitted which is again a spherical wave front moves away from this. Now both these the time difference between emission of the first maximum and the second maximum that time difference both are being measured by this frame of reference when I am measuring the frequency or time period in S frame of reference. And because these are being measured in S frame itself because that is the source which is emitting the light therefore when it is emitting the first maximum and the second maximum the time difference between these two is proper time difference because both these time that the time difference between these two is being measured only in S frame of reference. So therefore the T prime or the time period in S frame of reference is a proper frame of reference therefore if it is a proper time interval therefore an observer on the ground will find that the time interval between these two events emission of first maximum second maximum will be dilated because in that frame of reference when the first emission was there it started from here. The second emission according to S prime occurred when this particular source is moved away. So first maximum was emitted from one particular point the second maximum was emitted from different point. So these two events emission of first maximum emission of second even maximum according to S did not occur at the same point by that time the source has moved away but in the source frame of reference both of them occurred at the same place. Therefore time interval between these two emissions of maximum is proper in S frame of reference therefore in S frame of reference this will appear to be dilated. This is what I have written. Consider transfer stoplar effect the two consecutive maximum minimum in the field of electromagnetic wave are produced at same place in S. Hence the time interval between them which we call as time period is proper in S frame of reference should be S frame of reference and therefore in S frame it will be dilated. So this is what is the picture as I have said it is coming from here. So when first maximum was emitted and the second maximum was emitted by this particular person according to S he was at the same position but according to S when first was emitted it was here when second was emitted the demand has moved away. Therefore these two events did not occur at the same place. Therefore in S frame the time interval will be dilated. So using time dilation we can write that time interval or time period in between these two emissions in S frame will be equal to gamma times T prime as we know by wave theory this time period is inversive proportion or is equal to 1 divided by frequency. So this T I can write as 1 upon frequency this T prime I can write as 1 minus 1 divided by mu prime because this is a frequency in S frame. This T is also in S frame of reference which gives me exactly same as mu prime is equal to gamma nu exactly the same expression which we have earlier found out by using energy transformation. So what I am trying to say is that transverse Doppler effect can be thought as arising due to time dilation phenomenon and because classically there is no time dilation therefore classically we do not have any transverse Doppler effect. Let us take one or two examples and try to at the end of the lecture we try to give some idea of how the Doppler effect can be used. Let us first take a very simple example. The question is what is the required speed relative to an observer if a source of gamma ray of energy 14.4 keV when we say keV means kilo is equal to 1000 it means 14.4 multiplied by 1000 electron volt if its energy is to be increased by 10 to the power minus 6 electron volt just like you to appreciate the numbers this is 14.4 keV 14.4 multiplied by 1000 electron volt and this is 10 to the power minus 6 electron volt it is a very small change you know 9 orders of magnitude because this is of the order of 1000 electron volt and this is 10 to the power minus 6 electron volts I want to change so there is a gamma source source which is sending or emitting electromagnetic waves gamma rays of 14.4 keV and I want to change its frequency very very slightly of the order of 9 sort of magnitude 10 to the power minus 9 you know of the order 1 and 10 to the power minus 9 of that order I want to shift its energy by 10 to the power minus 6 electron volt then I want to apply give it a Doppler shift what is the speed that I require so that its frequency can be increased by 10 to the power minus 6 electron volt very small very tiny amount of course as we have seen the frequency has to be increased it means the source must move towards the observer therefore this particular source must come towards you person who is observing this frequency this frequency as I am sitting here I want to observe it increase by 10 to the power minus 6 electron volt therefore the source which is emitting this particular gamma rays must come towards me and as we will be seeing that you know this speed is definitely not very small in comparison to see what is the expression that we have just now derived for the case of v much smaller than c so this is what I have written because the energy has to be increased the source should move towards the observer it should not be the observer it should be observer then I use that particular expression h nu and that I have multiplied by h, h nu ds must be equal to h nu naught into 1 plus v by c here I have used plus expression because I want to increase the frequency that has to be taken as minus v so the sign has to be changed now I put the expression this h nu naught I take on the left hand side so this becomes h nu prime ds minus h nu naught should be equal to h nu naught multiplied by v by c this change I want 10 to the power minus 6 electron volt so I will put 10 to the power minus 6 here this I know as 14.4 keV v by c I can calculate the speed which is running out to be 2.08 centimeter per second these are very reasonable speed you know very easily can be manipulated in the laboratory can be observed in the laboratory therefore Doppler effect provides you a facility by which you can shift the energy of sources by very very small amount and therefore as we will be seeing just now it can be used for some very great studies which are generally not possible by any other means to resolve that type of small what you used to call in spectroscopy hyperfine field structure where the energy levels are differing very very by small amount that can be done by you know by Doppler shifting the source the gamma ray source in this case it is a gamma ray and 14.4 keV v by the way is a very standard source which people use very often when cobalt 57 decays 257 it emits a radiation of the 14.4 like kilo electron now this has certain implication and the two implications I would like to discuss today one is emission and absorption let us just go back to atomic physics and let us go to very very simple thing we do not want to talk it is not a course on atomical molecular physics we do not want to talk too much in details about the spectroscopy let us just assume that there could be something like hydrogen atom or something like that which we know very well when there is a transition takes place it emits a photon of course this concept of transitions is that came much later but nevertheless let us try to see the implication of this particular concept of photon and also the Doppler effect as far as these things are concerned suppose there is a transition then as a result of transition a photon is emitted we know that this is the way photon is emitted in atomic theory now because we know that photon has a momentum so strictly speaking whenever we are writing an expression for finding out the energy of the photon as a result of certain transition this momentum must also be conserved which essentially means that if we have let us say an atom which undergoes a transition and emits a photon and let us assume that this particular atom was at rest in a given frame of reference this photon has a momentum which is given by h nu by c because initial momentum was 0 therefore final momentum also has to be 0 it means this atom must require back if this atom requires back then this will also gain some energy some kinetic energy and if it gains some kinetic energy that kinetic energy also has to be supplied by the transition energy because once the transition has taken place that energy has to be partly utilized for creating the required energy and partly for giving energy to the photon it means the energy of the photon will turn out to be slightly smaller than the transition energy available to you this is what I have written a consequence of photon having a momentum is that emission and absorption would also obey momentum conservation this one has to realize let us consider the case of emission of a photon and let us assume that all the required speeds are non relativistic let us not consider a very high energy transitions at the moment therefore we can use partly the classical expressions of energy momentum kinetic energy and momentum relationship so let us suppose there is a transition which takes place from n state to m state energy corresponding to this particular level was E n corresponding to this level was E m so this is the total energy available to me as a result of transition this energy will be now used for two ways it will give energy to the photon and also it will give required energy to the atom therefore I must write E n minus E m is equal to h nu plus k r where k r is the kinetic energy of recoil and I must get a second equation which is the momentum conservation which is h nu by c is equal to momentum of recoil of course this momentum and kinetic energy in the classical case are related by a very simple expression k is equal to p square by 2 m and let us assume that m n is the mass of the atom which I am calling mass of nucleus or mass of the atom whatever you want to call it it depends on what time because general nucleus is the most heavy that is why I have written m n so p r can be written as under root 2 m n into k r now I can substitute this k r by writing this as p r square by 2 m n and therefore can find out an expression for the frequency of the emitted photon which I have written so this is the expression which actually is a quadratic in h nu here there is h nu square and in principle you have to solve this particular quadratic equation to get the correct expression of h nu but we must realize that this h nu will be slightly smaller than E n minus E m of course I must also say that generally if you take for example hydrogen atom case the energy that we are talking is of the order the ground state energy of hydrogen atom is minus 13.6 electron volt we are talking of tens of electron volt while m c square for the case of proton which is the nucleus which is one of the lightest nucleus nucleus is of the order of 940 MeV so 10 power 9 electron volt so this expression is extremely negligible therefore most of the time we neglect the coil especially when we are talking in the case of hydrogen atom but if we are talking of some nuclear transitions in that particular case that a coil can be significant and therefore we may we must account for this particular thing similarly if you are talking of absorption of photon if a photon has to absorb it has to not only supply energy for the atom to go to the higher energy state but this atom when this photon when it is absorbed the atom will start moving in front because momentum has to be conserved because initially there was a non zero momentum once photon is absorbed the same momentum must now be carried by an atom therefore the expression will become h nu is equal to E n minus E m plus k r exactly in the same fashion h nu upon c will be equal to p r and in this particular expression we will get E n minus E m is equal to h nu divided by 1 minus h nu upon 2 m and c by c square so you can see here there was a plus expression here there was a smaller expression the E n minus E m will be slightly larger than E n minus E m now my question is that what we call as a resonant absorption is it possible that we come across a situation when a photon is emitted as a result of a transition and the photon is used for causing similar inverse transition in some other atom it means a photon has been created when the atom has gone from nth state to mth state now this photon has come out now it is being absorbed by exactly similar atom and you want this particular atom to go from m state to n state is it possible we have just now said that in the first case the energy of the photon will be slightly smaller than the transition energy in the later case you require the energy to be slightly larger than E n minus E m see first-day quantum mechanics also gives you that there is all the photons have a particular line width you cannot have a perfect monochromatic photon so it so happens that if the line width turns out to be larger than the difference between the transition between these required energies of that order now we can sort of resonant absorption can still take place but in the case of many nuclear transitions it is not possible to do that so for example if this is your emission line which has this is your E naught which is E n minus E m and this absorption line they have finite line widths and this is because of the recoil the difference is because of the recoil and then this recoil energy if this is very large in comparison to this line widths hardly a resonant absorption will take place so that is where comes the Doppler effect and the recoil energy whatever is the recoil energy can be compensated by providing a slight amount of Doppler shift can I increase the energy can I or decrease the energy of the photon by giving the source a certain amount of energy because of the Doppler shift can it be shifted in 1950 Moon was a scientist who carried out some experiments where he could demonstrate the resonant absorption by imparting speeds to the gamma ray of the order of 7 into 10 power 4 centimeters per second remember we are talking of gamma rays which are much higher energies not the 13.6 electron mode that we have been talking earlier so in this particular case it required energy of the order of 7.10 to power 4 centimeters per second and then he could observe a resonant absorption in a situation in which it was not possible to do why this particular resonant absorption is interesting see what happened in a much later time Mossmauer observed that he can really in certain situation he can create I mean there are certain ways in which he can create a situation in which the total recoil is taken by the solid and total recoil can essentially be neglect can be made essentially require less and when he observed this particular thing he in fact got normal price for this particular thing he can see could see a require less emission then what is the why we are talking about require so much the idea is that in that particular case in the case of require less emission all these lines which are which have a hyperfine field structure which you know you find that the lines the energy differences are extremely small they can all be studied by slowly applying a Doppler shift giving energy of the order of millimeter per second or centimeter per second and therefore can we can scan all those energy level diagrams even though the energy source is I mean quite high energy and the gap between these energy levels is extremely fine but still I can measure by shifting the energy slowly varying the energy of the photon by very small amount by giving it a speed and therefore these photons can get Doppler shifted so Moswar was awarded 1961 Nobel Prize in physics and opened up possibilities where one can see small changes in the energy levels caused by the external effect using Doppler effect lastly speaking I will give quickly an example which is very interesting it is called laser cooling of a gas normally lasers are always thought of for hitting something normally used laser to for cutting something or belting something it is never thought that laser can also be used to cool something but it is interesting that laser can be used to cool gases let us see how see we realize that when we talk of a gas the temperature of the gas is related to the root mean square values of the speeds of the molecules if the root mean square speeds are high the temperature is high if they are smaller then the temperature is smaller so if by some means I can reduce the rms speed or root mean square speed of the gas molecules the gas will cool down now let us see how Doppler effect helps in this particular case let us in this particular situation take a very very specific case let us imagine that this particular thing is a source of laser which is emitting photon in this particular direction and there is a gas molecule which is moving towards it if it moves towards it and if I adjust in such a fashion that for a given transition the energy whatever is the energy required the energy of the photon is slightly smaller than that so the energy the photons which are being emitted have a slightly smaller energy than the transition energy then what will happen that this particular atom in its frame this particular photon is moving towards him this particular atom will find that the energy of this photon is slightly larger and if it becomes equal to the transition energy this photon will get absorbed by this particular atom and as a result of this absorption the photon of this the momentum of this photon will be transferred to this atom as this photon is moving in this direction this atom will slow down its momentum will go down now if an atom was moving not in this particular direction but was moving in this particular direction in that particular case it will not get absorbed because according to this particular person in this particular frame the energy of the photon will not equal to the transition energy which happens in this particular case therefore it will selectively absorb photons only which are coming towards it and therefore all the atoms which are moving in this particular direction their speeds are going to be slowed down now I want speed of the atoms to be slowed down not in one direction but in all the six directions what I do I create one source of these are photons here and this is generated by lasers another source here third source here fourth source here fifth on the top sixth at the bottom all moving towards gas so if a photon is moving in this particular I mean gas atom is moving in this particular direction its speed will slow down by the absorption from this photon if it is moving in this particular direction the speed of this will be reduced by absorption from this photon if it is moving in this particular direction the speed of this particular thing will get reduced because of the absorption from photon from this particular direction. And overall the speed of this can be reduced. So, let us just quickly read whatever I have written. Temperature is related to the RMS speed of the molecules. The speed can be decreased if a photon is absorbed moving in a direction opposite to that of gas atom. Let us take one dimensional motion and take the transition from E1 to E2 just a specific transition. Choose a photon source with h nu less than E2 minus E1. Only those atoms moving towards the source can absorb the photon causing the reduction in speed. This is what is called, the system is called, six lasers is called optical molasses. Creates a situation in all six directions. I must also say that when the photon is absorbed the atom also will emit it back, which we call as a spontaneous emission. But this spontaneous emission is generally random in all the direction. Therefore, it increases certain amount of temperature but we can show that overall the gas will cool down. If a true Kohenthenoji and Philip got Nobel Prize in 1997 for the laser cooling and related experiments. Now, I will summarize of whatever I have discussed in today's lecture. We discussed longitudinal and transverse Doppler effect as applied to the light and we gave some examples and its applications.