 So, let us start with from whatever we had discussed last time. We had discussed two examples of what in classical mechanics are known as inelastic collisions. In the classical mechanics we are used to not conserving the mechanical energy at that particular moment of time we used to just conserve momentum when such type of collisions were discussed. However, we had discussed in our last lecture that in relativity we have to also consider conservation of energy along with conservation of momentum. Hence, we had to apply conservation of momentum which eventually led to increase in the rest mass of the combined particle from the sum of the initial rest masses. So, if we have two particles which collide with it with respect to each other and then they get stuck to each other then if this is m1 this is m2 then the combined mass becomes larger than m1 plus m2. This is because we are conserving energy and we say that effectively this type of energy is actually never lost and whatever was the energy probably has been converted to some different form of energy not really mechanical energy. So, these are the type of examples that we discussed last time when we said that even in the classical inelastic collision we will have to conserve the total relativistic energy. Now, let us go ahead you know in our last lecture we had introduced a concept of zero rest mass particle. So, let us go little bit ahead with this concept of zero rest mass particle and try to discuss in today's lecture little more about this zero rest mass particle. See we had said at that particular time that it is possible that this particular particle could have a finite or non-zero energy and a non-zero momentum but in that case this particular particle must move with speed of light. And if it so happens you know that this is the standard expression for energy to momentum this is the total relativistic energy e square is equal to p square c square plus m0 square c to the power 4. So, if m0 becomes 0 this relationship just reduces to e is equal to pc it means for a particle with zero rest mass the energy and the momentum would be related by such simple equation as e is equal to pc. Of course, this particular particle must be travelling with the speed of light. Now, let us introduce the concept of photon it was originally in the explanation of photoelectric effect that the concept of photon was evolved. It was realized at that particular time that the type of frequency dependence of the energy of the electrons which are released as a result of photoelectric effect can be explained only if we assume that light is particle light. So, in fact we discuss in photoelectric effect experiment that light has a dual nature it could show as a particle it could also show as a wave. So, photon was essentially the idea of photon was created with a particle concept of light in mind and that was necessary to explain the photoelectric effect. And this particular particle was considered to be having energy of h nu where nu is the frequency of the wave that we have been talking about. So, if there is an electromagnetic wave or there is a light wave with frequency nu we can imagine that this particular light consists of photons each one of them has a energy h nu and of course, this particular photon travels with the speed of light. Therefore, this photon qualifies to be called a particle with zero rest mass. So, this is what I have written explanation of photoelectric effect indicated dual nature of light and light could be imagined as consisting of photons having energy h nu. But we have just now seen the relationship e is equal to p c. So, we can find out what will be the energy of the what will be the momentum of the photon. So, this is what I have written in the next transparency photon is treated as a particle with zero rest mass the energy and momentum can be related by this particular expression because e we have just now seen from the explanation of photoelectric effect experiment that e is to be given a value equal to h nu and p must be equal to e by c from the relativistic expression that we have earlier considered for a zero rest mass particle it means the momentum of the photon must be given by h nu by c. So, whatever we are saying about special theory of relativity is correct and the way we have invoked zero rest mass particle is correct in that particular case a particle like photon must have a momentum also which is equal to h nu by c. It was generally known even in earlier cases that light carries momentum, but here we sort of get involved with a rather newer concept when I treat particle photon just like any other particle but travelling with the speed of light and having energy h nu and a momentum of h nu by c. Actually in the photoelectric effect experiment it is not really possible to test the momentum of the photon because actually the photoelectric effect has to be observed when the electron is bound inside an atom and therefore, quite a bit of recoil energy is taken by the atom or by the solid and therefore, it is not possible to really test that the photon has a momentum. In fact, in photoelectric effect experiment we only apply conservation of energy. However, there is another experiment which we call as Compton effect in which it is possible to really check and experimentally verify that this idea of photon is a realistic idea as far as particle of zero rest mass is concerned it can really be treated with a particle energy equal to h nu and a momentum equal to h nu by c. So, let us just describe the Compton effect experiment. So, Compton effect is actually required of a photon by a free electron. So, let us assume that we have an electron E which is at rest and there is a photon which comes here with energy h nu this photon gets scattered and eventually comes out in a different direction. This direction makes an angle theta with the initial direction of the motion of the photon and it comes out with a different energy the changed energy h nu prime and as a result of this scattering this electron scatters goes other way at an angle of phi from the incident direction of the photon with an energy E. So, this is just like a simple classical scattering problem which we are quite used that one ball coming and another ball coming hitting and one ball going at an angle and another ball requiring except that the particles that we are talking here one is photon and another is electron. So, what we want to do we will treat photon as really a particle of energy h nu and a momentum of h nu by c and would like to apply conservation of energy and momentum to this problem like we have been applying for any other particle that we have discussed earlier. And let us see what result that we get. So, what I will do in the next transparency I will write the energy conservation and the momentum conservation because that is what we are supposed to apply. Now, as far as energy is concerned the initial energy before scattering was the energy of the photon it does not have a rest mass energy because m naught is 0. So, m naught c square is also 0. So, the total energy of the photon is h nu. This particular electron which was at rest has only the rest mass energy which was equal to m naught c square. So, total initial energy will be given by h nu plus m naught c square. After scattering you have a photon which goes this way and it comes out with the reduced energy or a different energy in principle it will be reduced energy because part of the energy will be taken by the electron. So, this particular h nu prime is the new energy of the photon change the energy of the photon. So, the photon which comes out has the energy of h nu prime and the electron now also gets certain amount of kinetic energy and let us assume that it is total relativistic energy is E E. So, the final energy after the scattering must be equal to h nu prime plus E E. So, this is what I will be calling as conservation of energy. So, I will get one equation for conservation of energy. I have to also apply conservation of momentum. When I apply conservation of momentum, momentum being a vector quantity I have to consider both x and y direction. I have shown this as the x direction which is the direction of the initial motion of the photon and perpendicular to it is the y direction and I have assumed that this scattering has taken place in x y plane. So, I have to conserve not only the x component of the momentum, but also the y component of the momentum. So, we realize that as far as this particular particle is concerned the photon is concerned its initial momentum was h nu by C we have just now discussed that. This electron was at rest. So, its momentum was 0. This particular photon has momentum only in the x direction. Therefore, the x direction initial momentum is h nu by C and the final momentum in the x direction will be the momentum of this component of the momentum of this photon along x direction and the component of the momentum of the electron along this particular direction. In the y direction which is this direction there was no initial momentum. So, final momentum should also be 0. It means if I take the component of this particular momentum along the y direction and component of this momentum of this electron along the y direction that too must balance out. So, these are the equations that I am going to write. Only thing I will write the momentum of the electron as P subscript P, P subscript E. So, let us now write all the three equations which I am putting in the next transparency. Let me just show here my picture to make it clear. This was my initial photon. This was my electron. This was going in this particular direction. Make an angle theta and this was electron which was going in this direction making an angle of phi. So, my first equation says which is the conservation of momentum along the x direction. This is the initial momentum of the photon because initial momentum was only along the x direction which is h nu by C. Now, momentum of the photon which has been scattered is h nu prime by C. This is h nu prime by C. So, if I take the component along the x direction, it becomes h nu prime by C cos theta. Similarly, the momentum of this electron let us assume is P E. So, along the x direction this component will be P E cos phi. So, this is what I have written in this particular transparency h nu by C is equal to h nu prime by C cos theta plus P E cos phi. Now, we take the momentum along the y direction. If I take momentum along the y direction, this momentum must cancel out with this particular momentum. The momentum of the photon, outgoing photon in this particular direction is h nu prime by C sin theta. Momentum of this electron is P E and in the y direction it will be P E sin phi. So, I expect that h nu prime by C sin theta must be equal to P E sin phi. So, this is the second equation which I have written here, h nu prime by C sin theta is equal to P E sin phi. And this is of course, we have discussed is the conservation of energy which is initial energy, Rasmus energy of the electron plus the initial energy of the photon is equal to outgoing energy of the photon plus the total relativistic energy of the electron which of course means Rasmus energy plus kinetic energy. Of course, we are using symbol P E here and E, they are not independent variables because energy is always related to momentum. Therefore, we have a further equation which relates energy to the momentum which is written as E E square is equal to P square C square plus M naught square C to the power 4 standard equation with standard relationship between energy and momentum. Now, we have to work out with these equations and try to find out what will be the scattered frequency or energy of the photon or energy of the scattered photon. As we will be discussing little later in the experiment, it is not always very easy to know the parameters of the electron that is scattered. It is somewhat easy experimentally to calculate the energy of the photon whether it is incoming photon or whether it is outgoing photon. So, what we will try to do out of these equations, we will try to eliminate those things which are having relationship with the electron. So, first thing that I would like to relate, remember phi was the angle in which this electron got scattered. So, there is a cos phi here, there is a sin phi here. So, I will first like to eliminate this phi which is rather easy to do it because this P e cos phi, I can take this particular thing to the left hand side. So, this right hand side will become just P e cos phi. Here on the right hand side is just P e sin phi, I square and add of them, add the two equations. If I do that, I will get here P e square cos square phi plus P e square sin square phi because cos square phi plus sin square phi is equal to 1. So, I will just get P e square. So, that is the equation which I am writing in the next transparency. So, I am writing P e square is equal to cos square phi plus sin square phi which should be equal to P e square. This term was a term which I have taken in the first equation from the right hand side to the left hand side. Let me just go back. This is h nu by C and this was h nu prime by C cos theta. So, I have taken this to the left hand side. This is what is written here, h nu by C minus h nu prime by C cos theta. And of course, I have squared it because I have squared and added it. In the second equation on the left hand side, we had just this term h nu prime by C sin theta. So, I have squared and added up. So, this is what I have got after using the first equations and eliminating phi. Now, just to make these equations look somewhat simpler, what I have done? I have multiplied by C square from the both the sides. So, if I multiply by C square here, I will get P e square C square which appears here because I have multiplied by C square and there is already a square here. So, inside it will get multiplied by just C. So, when I multiply this by C, I will just get h nu here. When I multiply it by C, I will just get h nu prime here. When I multiply by C here, I will get h nu prime. So, I get this equation as P e square C square is equal to h nu minus h nu prime cos theta square plus h nu prime sin theta square. Let us open this equation and try to see whether we can still simplify it. So, I have to just take the squares and again try to manipulate the terms. This is what I have done here. Same equation which I have written on the last transparency, h nu minus h nu prime cos theta square plus h nu prime sin theta square. So, I first expand this term. So, I will get a square plus b square minus 2 a b. So, a square is h nu square which I put h nu square. b square is h nu prime cos theta square. So, it becomes h nu prime square cos square theta minus 2 a b. So, minus 2 into h nu into h nu prime cos theta. So, this becomes the expansion of the first term. The second term is just h nu prime square sin square theta, which I have kept just like that. We realize here that I have h nu prime square cos square theta and there is h nu prime square sin square theta. So, when I take these two terms into consideration together, I will get h nu prime square multiplied by sin square theta plus cos square theta, which again gives me one. So, these two terms together will just lead me h nu prime square. So, I go to the next step. This h nu square I just put it as h nu square. These two terms gives me h nu prime square. So, this I put as h nu prime square. This term I keep it as it is, which is minus 2 h nu into h nu prime cos theta. So, this is what I have got after elimination of the phi term in relationship between energy, sorry momentum and this particular theta corresponding to the energies of the incident photon and the outgoing photon. Now, this p square c square I can also evaluate from conservation of energy and as I said our idea is to eventually eliminate anything which has to do with electron. Here we have the momentum of the electron. Here on the right hand side, there is nothing which is relating to electron. This is relating to the incident photon. This relates to the outgoing photon. Similarly, these two terms. So, I would like to get eventually rid of this particular term p e also and there is easy way because I know that e square is equal to p square c square plus m naught square c to the power 4. So, what I will write? I will write this p square c square in terms of the energy of the electron. So, that is what I am doing in the next transparency. So, this was the energy and momentum relationship for the electron. I have just taken p e square c square on the left hand side and written this as e e square minus m naught square c to the power 4. Now, this energy I have taken from the energy conservation which I have written earlier. This was my energy conservation equation m naught c square plus h nu is equal to h nu prime plus e e. So, from this I can extract e e. This e e will be m naught c square plus h nu minus h nu prime. So, I know what is e. So, I will substitute this e e here. Then I will get on the right hand side everything in terms of h nu. This p e square c square I will eliminate from the first equation which I just now written. And therefore, I will get rid of energy of the electron as well as the momentum of the electron. So, let me write this particular equation which I have written here. For e as we have just now seen, I have written from this equation h nu plus m naught c square minus h nu prime. I just slightly reorganize these terms minus m naught square c to the power 4 which was here which I just brought here. Now, expand this particular equation. Now, we have a plus b plus c square. So, we have to take a square plus b square plus c square plus 2 a b plus 2 b c plus 2 c a all those things. So, let us do it. So, let me just take the square of this particular quantity. I will take h nu square which is here. I take the square of this quantity which is h nu prime square which is here. Remember there will be a positive sign here. Then there will be a square of m naught c square which becomes m naught square c to the power 4. And this m naught square c to the power 4 will cancel with this m naught square c to the power 4. So, I am not writing neither this term nor that term. So, I have written only these two terms out of the square terms. Now, I have to write 2 a b. When I write 2 a b, I write this as 2 h nu into h nu prime to m naught c square which I have written here. 2 h nu into h nu prime this a and this is b. So, 2 h nu into h nu prime because there is a negative sign here. So, this becomes minus 2 h nu into h nu prime. Now, this m naught c square has to be multiplied by this with a factor of 2. This also has to be multiplied by this and have a factor of 2. So, I have combined these two terms and have written here as plus 2 h nu minus h nu prime taking them together multiplied by m naught c square. So, I have just expanded this particular term here and using this particular thing I get this as the value of p e square c square. Let us see whether we can simplify it further. This is what I have written earlier h nu square plus h nu prime square plus 2 h nu minus h nu prime m naught c square minus 2 h nu h nu prime. This I am equating with p e square c square which I had obtained earlier for elimination from the conservation of momentum equation which if you remember was h nu square plus h nu prime square minus 2 h nu h nu prime cos theta. Let me just bring you back to that transparency. This is what we have written earlier p e square c square is called h nu square plus h nu prime square minus 2 h nu h nu prime cos theta. This equation was obtained by eliminating phi from the first two equations of the momentum conservation. The second p e square c square I have got by applying energy conservation. So, I can equate this right hand side to that right hand side which I have done in this particular transparency. So, h nu prime h nu square plus h nu prime square minus 2 h nu h nu prime cos theta from momentum conservation. This is from the energy conservation which I have written. You can see very easily that I can simplify these equations further because h nu square will cancel with this h nu square this h nu prime square will cancel with this h nu prime square. Here what I will be getting this two factor will also cancel and if I slightly organize bring this particular term on the left hand side what you have 2 is already cancelled here. So, you will get h nu into h nu prime. This being on the left hand side will give me 1 minus cos theta on the left hand side and only this term on the right hand side which is h nu minus h nu prime m naught c square. So, this is an equation which must be obeyed if whatever I am saying is correct which gives me the relationship between the frequency or the energy of the outgoing photon in terms of the energy or frequency of the incoming photon and the angle of scattering which is theta. This equation can be further simplified if we write in terms of lambda. Let us try to work out this particular example in a slightly simpler fashion. This is what I have written earlier the last transparency. Now I express everything in terms of lambda. The frequency of the incoming photon can always be written as c divided by lambda because frequency multiplied by lambda must be equal to speed and the speed of photon is c. Therefore, I can write nu s c by lambda which is the wavelength of the incoming photon. h nu prime I can write again as h c by lambda prime because nu prime is equal to c by lambda prime. So, this is what I have written as c by lambda prime. Remember in general we will find that nu prime has to h nu prime has to be smaller. It means lambda prime has to be larger into 1 minus cos theta which is the same term. Here also I express nu s c by lambda and nu prime s c by lambda prime writing this equation in this particular fashion. So, this h nu has been written as s c by lambda. This h nu prime has been written as s c by lambda prime. So, I can cancel lot of h c's. You can see there is an h c here, there is an h c here, but there are 2 h c's here. So, probably one of them will remain, will cancel one of the h c's. There is another c, here there is m naught c square. So, there will be one c which will be going away and the equation will turn out to be fairly simple. This was my equation which I had written here earlier. So, I get h c square because there are 2 h c's here. This is lambda into lambda prime which I have written here. Here I take h c. I just use this particular thing, simplify this, taking h c out. So, it will become lambda prime minus lambda. So, this is lambda prime minus lambda and lambda lambda prime in denominator which I have written in denominator multiplied by m naught c square which is this equation. Now, I start cancelling the term. I write this on the left hand side now, lambda prime minus lambda, this on the right hand side. This h c will cancel with this h c. So, there will be one h remaining here and one c remaining here. There will be m naught c square. So, one of the c will cancel with this c. So, there will be one h remaining here and on the right hand side there will be just a m naught c which will be remaining here. So, the equation which I get is as follows. Just like trying to write in a different fashion, trying to write on the right hand side, I get the standard well-known Compton effective equation which says lambda prime minus lambda is equal to h upon m naught c multiplied by 1 minus cos theta. It means whatever I am trying to say, the scattered photon will have an increased lambda and that difference between the original lambda and the increased lambda will be given by h upon m naught c 1 minus cos theta. Cos theta being always smaller or at the most equal to 1, either lambda prime minus lambda will be equal to 0 or will always be positive as I had expected because part of the energy of the photon must have gone to the electron. Therefore, the photon which is coming out must come out with somewhat reduced energy. It means reduced frequency and therefore, larger lambda. So, this is the standard Compton fact expression. Now, question is that can I experimentally verify it? If I really experimentally verify it, then I have sort of shown that my idea of photon as a zero-res mass particle with energy ash nu and momentum ash nu by c has a meaning and probably it is something which we can take forward. Now, it is very easy to say that you have a free electron and a photon comes and hits here, but where do I get when I want to perform an experiment, a free electron? It is not easy to get a free electron and perform this particular experiment when you want to do really experiment. So, generally in a Compton effect experiment, we use a metallic foil. When I use a metallic material, in metals, they are supposed to be very large number of nearly free electrons. I mean, I mean, strictly speaking, they are not really free, but to a reasonable approximation, I can treat them as more or less free. The other problem is that these electrons which are somewhat free, they may not all be at rest, but that does not matter. Let us, if you have to perform an experiment, we have to see that. So, what we do? We take a metal and allow frequency electromagnetic wave. In fact, as you will be seeing that this H upon M naught C turns out to be extremely small. Therefore, the change in the lambdas are very small. It is order of one tenth of an angstrom, which is very small. So, normally when we perform this experiment, we use gamma rays. So, this is the way we perform the experiment. We use free electrons within the metals for the experiment. And experiment is something like here. You have a metallic target. There is incident gamma ray beam which comes and hits here. And you have a detector which detects gamma rays. It is possible to obtain detectors which can detect and which can count, which can tell how many photons have come at a particular angle. So, if my photon beam is initially coming in this particular direction and I am collecting my photons in this particular direction by counting the number of photons reaching detector and I can measure their frequency which is also easy to measure, then I know all these photons must have come at an after scattering at an angle theta. Of course, this theta I can change and keep on verifying my experiment. So, this is what normally is done that you allow this particular gamma ray to be incident on metallic target. You have detector, fix at a particular theta, collect the photons, measure their energy, keep on changing theta, keep on going to different thetas. And at every angle, you keep on measuring the number of photons and keep on finding out their energies. When we do this particular experiment, we get something like this. The curve is somewhat like this. So, if we take the lambda prime which is the frequency that wavelength that we measure and find out how many number of photons are coming with that particular wavelength approximately. Now, in fact, you have to take a range between lambda and lambda plus d lambda. But let us not go into those details. We just take n versus lambda prime curve. You find two peaks. One at larger lambda, another at smaller lambda. This peak is generally having much larger width. Now, this original peak corresponds essentially to the original wavelength lambda which was present there. This particular maximum does indeed corresponding to the lambda prime that is expected out of the Compton scattering. So, you do expect, you do find experimental verification. Of course, there is a line width. I mean, we know that there is a natural line width which has to be always present. But here the line width seems to be broadened especially in this particular case. Also, we find that the original wavelength is also present. Now, this can be explained as follows. When the photons are coming and being incident on the atoms of the metal, not every electron, sorry, not every photon will get scattered only from the free electron. Some of these photons will also get scattered from the atoms. And if they get scattered from the atom, remember in the Compton effect experiment what was appearing was H upon M naught C. So, this M naught for the atom is going to be very, very large. In fact, you know, had it not been electron, but any other particle, this expression would still be alright, except that mass has to be different. Mass that you have to use is has to be of the mass of the particle that is scattering photon. And if that happens to be an atom, this M naught is going to be extremely large. Therefore, lambda prime minus lambda is going to be extremely small, which is essentially negligible. Therefore, you will not find any measurable change if the photon gets scattered from the atom. Therefore, you do also find the original lambda even though you are measuring or you are collecting photons at an angle theta, because the original wavelength is also present, because these are the photons which have got scattered from the atom and not just from the electron. Also, we see a larger line width and this particular line width is expected because not electrons are not really at rest. They are moving with different type of velocities and therefore, there has to be some amount of distribution of the photon energy that you are measuring. Therefore, this particular width is being explained because of larger number, a distribution of velocities of the electrons that you see in the metal. But basic crux of the issue is that you do really see that this particular experiment is just does prove that it is possible to get lambda prime minus lambda is equal to h upon M naught C 1 minus cos theta. This particular peak does satisfy that equation. Therefore, we can treat photon really as a particle with energy as nu and a momentum as nu prime as C. So, this I have sort of written in the conclusion of the Compton effect experiment. The original wavelength is due to the scattering from core and large line width due to velocity distribution of electrons. Now, once we have come to these issues of photon, let us try to discuss one or two problems based on the photon. All that we have said now that photon is like in relativity is like any other particle. Only thing special about the photon is that its rest mass is 0. Otherwise, as far as conservation of energy and momentum are concerned, they can be treated just like any other particle. So, let us take one particular example in which there are particles involving or also involving photons. So, this is my example. Say example is little bigger. So, let us try to slowly understand this particular example before we try to work it out. So, there is one particular particle which has a rest mass of 135 MeV by C square. See, as I have told you earlier that rest mass is often expressed in the units of energy. So, we can write this as 135 MeV, but actually it means is that rest mass is 135 MeV by C square. It is M naught C square which is 135 MeV. So, we will write in this particular expression the energy always in the unit of MeV by C square. So, it has a rest mass of 135 MeV by C square. I am sorry, I will write mass in the units of MeV by C square and is found to be 0.8 C. The speed of this particular particle is 0.8 C in a frame S prime. Normally, we have been giving in the frame S, but here problem is slightly different. So, this particular speed has been given in a different frame S prime and the speed of this particular particle as measured in S prime is 0.8 C and this S prime itself moves relative to S with the speed of 0.6 C. So, we have a frame S. In this frame, S prime moves with the speed of 0.8 C, I am sorry, S prime moves with the speed of 0.6 C and in this particular frame S prime frame, a particle moves with a speed of 0.8 C. Now, next part of the problem, all the motions are collinear and are taken to be along x direction. So, we assume that everything is moving in one line. In S prime, it is found that the particle decays into two photons. So, what is found that of course, if it decays in S prime, it has also to decay in S, but the final things are all given in the frame S prime. So, what we find that this particular particle decays into two photons. So, particle is no longer existing, but it has now become two photons. Two photons have emerged and the particle disappears, so to say. Now, it is found that in this particular frame of reference S prime, each particle makes an angle theta prime with the initial direction of the particle motion. Now, question is, find the angle theta prime and the frequency of photon in this particular frame. In fact, instead of frequency, I can just find the energy. Once I know the energy, I can always find out the frequency. Find the angle theta and the frequency also in the S frame. Though my experiment has been described in S prime frame, but I have also to find out these corresponding quantities in a different frame S frame. So, this is my picture which I have shown here. This was my S frame. This was my S prime frame. S prime moves relative to S with a speed of 0.6 c. I have shown everything in blue here just to match with this particular frame of reference. So, as far as the observer sitting in S prime is concerned, that particular observer observes that this particular particle suddenly disappears and two photons come and they make equal angle with respect to the initial direction of the motion of the particle. And this angle is theta prime. This is theta prime. This is theta prime with respect to the original direction of the motion of the particle. Let us assume that this photon has energy of E prime p1. Let us assume that this particular photon has energy of E prime p2. The question is, what is the frequency of the photon and other photons? And what is this angle theta prime? What has been given in this particular problem is only that this angle and this angle are same. Having now discussed conservation of energy and momentum and discussed the fact that photon is to be treated like any other particle, this problem is like any other simple problem of collision or scattering in which all I have to do is to apply conservation of energy and momentum, but relativistic equations have to be used. That is all. So, because the problem has been given in S prime. So, let us first try to find out before collision or before this particular decay has occurred. Let us not use the word collision. The decay has occurred. What was the energy and what was the momentum? So, everything is being written in the S prime frame of reference. Once I write in S prime frame of reference, the particle speed is 0.8 C. So, if I write S prime frame, the speed of the particle is 0.8 C. Using this particular U, I can calculate what will be the value of gamma U. And as we have seen earlier, gamma U corresponding to 0.8 C always gives out to be a comparatively cleaner number, which is 5 by 3. Once I know gamma U, I can find out what is the energy of this particular particle, because energy of the particle E is given by gamma U, M naught C square and M naught C square, I have been given is 135 MeV. Similarly, I can find out what is the momentum of the particle, which happens to be only along the x direction. So, I am not putting the vector sign is equal to gamma U M naught. So, this is what I have written in this particular transparency. Of course, because this S frame of reference, so I must use E prime is here also I must use U prime to be more precise, because everything is being observed and here also should be gamma U prime, because everything is being observed in the S frame of reference. So, let us look at this particular transparency. I have written gamma U prime is equal to 5 by 3 using value of U prime as 0.8 C. E prime is equal to gamma U prime M naught C square, which is 5 by 3 into 135, which is equal to 225 MeV, if you just calculate this number. P prime is gamma U prime, M naught U prime, 5 by 3. This particular value of M naught U prime, I have this particular value of U prime is 0.8 C. For M naught, I have used the value of 135 by C square, because M naught C square was 135 MeV. Now, what I have done here, I would like to express the momentum in the unit of MeV by C. So, one of the C I have retained here, one C cancels from this. So, there has to be one C in the denominator, because there is a C square here and there is a C. So, I cancel only one of the C, another C I have retained in the units. So, I have to just calculate this 5 by 3, multiplied by 135, multiplied by 0.8 and this C I am just taking as a part of the unit. So, if I calculate this number, I get 180 MeV by C. So, basically I have used these equations, which I have written here in this particular paper. Of course, this is M U prime, M U prime, gamma U prime, substituting these values, only thing expressing this particular quantity in the units of MeV by C. So, this is what happens in S prime before this decay occurred. Now, let us see what happens after the decay occurs. After the decay occurs, then what you are getting are two photons, one going this way, another going this way, making an angle theta prime with respect to each other. So, like before, like in the Compton effect experiment, I have to conserve initial momentum to the final x component of the momentum, I have to balance the y component of the momentum. That is what I am doing in the next transparency. If E prime P 1 and E prime P 2 are the energies of the two photons, the sum of the two energies must be equal to the initial energy that is available to me, that we have just now seen is 225 MeV. So, this is equal to 225 MeV. Along the x direction, I have to take, this is the energy. So, momentum will be this divided by C, because this is a photon, which is a zero brass mass. This is for the first photon. X component must be cos theta, just like Compton effect experiment, cos theta prime rather. For the second photon, you must have E prime P 2 divided by C cos theta prime. This must be equal to the initial momentum, because the initial momentum was only along the x direction, which was 180 MeV by C. As you can see here, it was 180 MeV by C. So, I am writing this as 180 MeV by C. In the y component of momentum, in the y direction, there was no momentum initially. So, final momentum must also cancel out. Therefore, E prime P 1 by C sin theta prime must be equal to E prime P 2 by C sin theta prime, because these thetas are same. Sin thetas have to be same. C, of course, have to be same. So, I get E prime P 1 is equal to E prime P 2. It means both the photons must have the same energy. And if they have the same energy, I know their sum is 225 MeV. So, the energy of each of the photon will be 225 divided by 2. So, I get immediately the energy of the individual photons. Now, I know the energy of the photon, which I can substitute in this particular expression. I can find out what is theta, because this theta is same. These two are same. So, I will get 2 into whatever this energy by C cos theta is equal to 180 MeV by C. So, I can immediately find out what is cos theta, which I am giving in the next transparency. So, as I said that E prime P 1 must be equal to E prime P 2, which is 225 divided by 2, which turns out to be equal to 112.5 MeV. And cos theta prime, in the right hand side, there was 180. See, this is equal to 225 by 2, but the two terms are equal. So, this becomes just 225. So, cos theta prime, this C cancels with this C. So, your cos theta prime just becomes 180 divided by 225, which is written here, from which you get theta prime is equal to 36.87 degree. So, approximately at an angle of 37 degrees, these photons will get scattered from the incident direction of the motion of the particle. Now, the question is that we have to find the same information in S frame of reference. What I can do? I have to use energy momentum transformation. I go back to S frame, find out the energy of the particle and do exactly the same thing as I have done here. So, that is what I am doing in the next transparency. I go back to S frame from S frame of reference. So, in S, transform the energy momentum of the particle before decay. And because I am going from S to S, I will be using inverse transformation. That is what is being shown in the next transparency. If you remember, this was the equation of the momentum transformation and I have put a plus sign because I am talking of inverse transformation going from S frame to S frame. So, P will be equal to gamma P prime plus V E prime by C square. So, I know what is P prime of the particle before a decay. I know what is the energy of the particle before a decay. I substitute in this particular expression. I should be able to get the momentum of the particle before it decays in S frame. Of course, gamma that I have to use is between S and S frame and their relative velocity is 0.6 C and not 0.8 C and corresponding to 0.6 C, we have seen many times earlier gamma turns out to be equal to 1.25. So, this gamma is 1.25. Momentum P prime was 180 MeV by C. So, this unit I have retaining here plus V which is 0.6 C. So, there is a 0.6 and there has to be one C will cancel with this C, another C will come out in this unit, this is 225 MeV. So, this everything becomes in the unit of 225 MeV by C. So, if I calculate this thing, I get 393.75 MeV by C. So, this is the initial momentum of the photon, sorry momentum of the original particle before it decays in S frame. Similarly, I can calculate the energy. For energy, I have to use energy transformation. Again, I have using a plus sign because I am talking of inverse transformation. So, E will be equal to gamma E prime plus V P prime, gamma is 1.25, E prime was 225, P prime was 185 MeV by C and there was a 0.6 C which is the speed. So, this C will cancel with this C, everything becomes in the unit of MeV. So, I get 1.25 multiplied by 225 plus 0.6 multiplied by 180 in the unit of MeV. So, my energy, if you calculate this number, turns out to be equal to 416.25 MeV. So, now problem is exactly similar of what we have discussed earlier, except that now the initial particle energy is 416.125 MeV and its initial momentum is 393.75 MeV by C. Then we apply exactly the same way transformation and conservation of energy and momentum and get the new angle, get the angle in S frame. So, that is what is being done here. It is just exactly the same equations that I had written in S frame except that instead of prime, I have removed those primes and the numbers that I am using are those numbers which correspond to the energy and momentum in S frame. As far as energy is concerned, S frame is 416.25. So, this I have written here. As far as the momentum is concerned is 393.75, which I have written here. I am not using prime because this is everything is in S frame and otherwise all the equations are exactly identical. Once I write this particular equations, exactly the same way I will solve, I will get E p1 is equal to E p2 and therefore, the energy will turn out to be half of the energy. I substitute it back and get the new cos theta. So, I get just E p1 is equal to E p2. That number divided by 2, we get 208.125 MeV and cos theta turns out to be 393.75 divided by 416.25, which is equal to 18.92. Now, we can verify these numbers. In fact, what we have done, we have used transformation before the particle got decayed. I could have even used the transformation after I have obtained the photons because their energy will also be transformed, their energy and momentum will also be transformed when I am going from S frame because for us, photon is like any other particle. So, I can use exactly the same transformation equation even for photon. So, that is what I am going to do next and show that I get the same result. So, I said verify, we can verify the results by transforming the photon energies and also momentum. So, let us look at photon 1 in S frame of reference. This is what I had got its energy 112.5. This is what we had calculated earlier. Now, let us calculate its momentum. This particular particle was moving at an angle theta prime from the incident direction. So, if I take the x component of the momentum, this was the total momentum of the photon multiplied by cos theta prime, which I had already calculated as 180 divided by 225. This was the energy, this C I take in the unit of the momentum MeV by C. So, the x component of the momentum, if you calculate this number, turns out to be 90 MeV by C. The y component of the momentum is this momentum multiplied by sin theta prime. If I do that particular thing, I take this particular value is 112.5 and take sin theta prime, this turns out to be 0.6 and I get this number as 67.5 MeV by C. So, I have found out what is the x component of the momentum, what is the y component of the momentum. Now, transform these things back to S frame. So, I calculate P1x, which is the momentum of this photon, x component of the momentum of this photon in S frame of reference. You see in transformation equation, there is no difference photon is just like any other particle. So, I write gamma times P prime for the first particle x component plus V for the first particle divided by C square. Gamma has to be 1.25, because I am transforming from S prime to S, exactly similar equations. This momentum is 90.6 multiplied by V is 0.6 multiplied by 112.5. You calculate this number, you get 196.875 MeV by C. Now, as far as y component of momentum is concerned, there is no problem, because they are same. So, I just write P1y is equal to P prime P1y, which is 67.5 MeV by C. Of course, I have taken the photon, which has a positive component in the y direction. If we take the other photon, that will have exactly the same value, but will have a negative component in the y direction if the photon was coming down rather than going up. So, otherwise this particular expression will be valid, this number is valid for both the photons. Now, I transform the energy of the photon. If I transform the energy of the photon, I get energy of the photon in S frame as gamma multiplied by the energy of the photon in S frame plus V multiplied by the x component of the momentum of the photon in S frame of reference. So, this gamma as we have seen is corresponding to value of 0.6 C, which is the speed between S and S frame. I substitute here 1.25. E prime P1, you know is 112.5. This V is 0.6 and this was 90. This is 0.6 C and this P prime was in the unit of 90 MeV by C. So, this C cancels out. I get everything in the unit of MeV. So, if I calculate the energy of the photon in S frame of reference, this will turn out to be 208.125. Now, I have to find out the angle in S frame that I can find out because I know the y component of the momentum and I know the x component of the momentum which is here. So, that is what I am doing in the this particular transparency. This is the value of the x component of the momentum. This is the value of the y component of the momentum. I just take 10 theta. This divided by this, this gives me the value of theta, which I get exactly same as 18.92. Of course, I have also found out the energy of the photon in S frame and what has been asking the problem is the frequency, but as I have said that if you know the energy, I can always find out the frequency by dividing it by h. So, essentially the problem gets solved. The basic idea of this particular problem is just to tell that like we are applying energy momentum transformation for any other particle, I can apply also for the case of the photon. So, in the end, I will summarize whatever we have discussed. We had introduced the concept of photon as a particle with zero rest mass. We gave an example and saw that photon momentum and energy conservation have how they have to be applied. Thank you.