 In our last lecture, we had discussed the Doppler effect in light. As we had said that Doppler effect in sound is well known, we visualize many times in our daily life. But in light requires slightly different treatment and probably not so much apparently known. We also discussed a special Doppler effect which we call as a transverse Doppler effect which is normally which does not have any classical analog will not be seen in sound for that matter. These two aspects we discussed then we gave some flavor of some experiments in which Doppler effect is used just to give an idea that you know the Doppler effect that we have talked has certain usefulness as far as certain part of physics is concerned. In fact it is even in medical sciences that Doppler effect is used quite effectively. So I think Doppler effect is finding more and more uses but what we did in last lecture just to give some idea about the Doppler effect. Today we would like to go to a slightly different concept basically it is for problem solving. We define what we call as a center of mass frame of reference. Normally in classical mechanics when we are not dealing with relativity the center of mass is always described in a fashion such that we have a standard definition of center of mass how to find it out and if we change our frame of reference to that center of mass of a system of particles then we call that particular frame of reference as center of mass frame of reference. The advantage of dealing with center of mass frame of reference is that if there is no external force to the system center of mass becomes an automatic inertial frame of reference because if there is no external force to a system of particles the center of mass must be addressed or must be moving with constant velocity as seen from any other inertial frame of reference. So if we move our self to the center of mass automatically we are landing up ourselves into an inertial frame of reference. Now we slightly modify this particular definition as far as relativity is concerned and we say that center of mass frame of reference is that frame of reference in which the sum of the momenta of all the particles is 0. This center of mass frame of reference is occasionally referred as C frame. So what we have said that center of mass frame is a frame in which the sum of momenta of a system of particle is 0. Normally as we have always said that there is a set of particles which we consider as a system in which we are interested in. See there could be forces in between the particles. It means a force being applied by one particle on any other particle that we call as the internal force and if a force is applied by a particle which is not a part of the system that is what we call as an external force. So I repeat that if we find that sum of the moment, if we go to a frame of reference in which the sum of the momenta of all the particles of the system is 0 that is what we call as C frame or center of mass frame. Similarly we can define or we generally call a L frame as the laboratory frame of reference. This frame of reference is a frame in which you have described your problem or you have described your experiment or in which you may perform the experiment. This frame of reference could be a C frame, could be a center of mass frame of reference or could be any other inertial frame of reference. So I may describe in a problem, my problem in a frame of reference which does not happen to be center of mass frame of reference or I could choose to describe the problem only in a center of mass frame of reference. So laboratory frame of reference is a generic frame of reference in which we have described my problem or my experiment. Now why we talk specifically of C frame especially in relativity we will see that many problems become very simple if we work it out in the C frame. Though the problem may not have been given to you in C frame, it might have been given in some other frame but still if you transform the problem into C frame the problem turns out to be simpler. So in today's lecture we will be essentially discussing some problems in which you will find the use of C frame to be convenient frame and probably in easy frame. So this is what I have written in SDR which I mean special theory of relativity problems can be simplified when transformed to C frame. I would also like to mention that when we actually perform many HEP, HEP is stands for high energy physics. If we perform experiment like collision experiments or scattering experiments in high energy physics many times we prefer to design this experiment in C frame of reference in center mass frame of reference. So that overall there is a energy which is same we will be seeing little later that you know energy turns out to be smaller in the C frame of reference basically because you do not have to spend energy to accelerate the center mass frame of reference. It effectively means that if you have one particular particle and you want to design an experiment in which another particle comes and collides it. In that case this frame of reference is not a center mass frame of reference because the net momentum is not 0. So you would rather like to have an experiment in which both the particles come together and collide such that the center of mass so that it becomes in a center mass frame of reference or the net initial momentum is 0. So let us come to one particular problem which is actually a simple problem but this illustrates the fact that how it can become simple if it is solved in C frame and it is rather difficult to solve in L frame even though the problem appears to be fairly simple. Let us look at the problem. There is a particle of Rasmus capital M naught which has a total energy of 3 M naught C square. Now we must be clear that when we are talking of total relativistic energy it means it includes Rasmus energy. So the total energy which means kinetic energy plus Rasmus energy is 3 M naught C square. This particular particle decays into two identical particles. So this particular particle goes away and actually decays into two particles. Each one of them have a Rasmus of M naught. So they are identical particles. Whatever might be the particles this may not be a realistic sort of experiment just for sort of understanding of our problem. So it decays into two masses, two particles both of which have a Rasmus of M naught. Now I have to find out what will be the velocity of these particular particles when they have decayed given the fact that the decay product it means both the particles move along the direction of the motion of the present of the parent particle. Let me just read again. A particle of Rasmus capital M naught and total energy 3 M naught C square decays into two identical particles of Rasmus small M naught. Find the velocities of the two particles given that the decay products move along the direction of the motion of the parent particle. If I want to draw into a small simple picture essentially it means that this is a particle which is of Rasmus capital M naught. This is moving in a given frame and eventually it decays into two particles, two small particles. Each one of them has a Rasmus M naught obviously because it has to conserve energy and conserve momentum therefore these particular particles will also be moving. What has been told that both these particles move in the same direction as this particular particle. So it means they also move in this direction, this direction or opposite direction but the total motion is confined only to one dimension which we can call for example x direction that does not matter what direction but everything the problem is just a single dimension problem a one dimensional problem in which everything has been described. On a traditional classical mechanics problem where we have to conserve energy and momentum this problem would have been absolutely simple but we will see that if we try to work out this problem in this particular fashion you will see that there are certain complications and it is not easy to work out this particular problem until we go to a different frame of reference which is the center of mass frame of reference. Let us look at this particular problem. First let us attempt to solve it only in the L frame the laboratory frame the way it has been described. We will discuss what are the difficulties which comes across. So as we said the idea is that we have to apply conservation of energy and conservation of momentum in this particular frame. Energy has been given because there is only one particular particle initially. So energy has been given as 3 M naught C square. So I have to find out this energy and we have to find out correspondingly what is the momentum of this particular particle. Once we find out the momentum I know the initial energy, I know the initial momentum. In the case of two particles we have to write initial energy is equal to the sum of the energy of the resultant particles. Similarly initial momentum vector is equal to the initial final momentum. That is what we have to do. So let us first find out the initial momentum. We have been given that E is equal to 3 M naught C square and if you remember what we have said that the energy is given by gamma M naught C square where M naught is the rest mass of any particular particle and in this particular case because the energy has been given as 3 M naught C square it is quite clear that gamma is equal to 3. Only thing because this has been described in laboratory frame of reference. So let us write a subscript L to make it clear that this gamma that we are talking or the velocity that I will be talking on which this particular gamma depends is actually the velocity in the L frame or in the laboratory. Now once we have this particular gamma L I can always find out what is the velocity of this particular particle and once I find out the velocity of the particle I can always find out the momentum of the particle. Of course I can also use that energy momentum relationship that also I will use just to show that we get the same result. Let us first try to find out the velocity value using this gamma L. If you remember the value of gamma the expression for gamma is given as 1 upon under root 1 minus V square by C square sometimes we have used V sometimes we have used U but I think it should be clear by now when we use V when we use U. Now I want to solve it for V upon C I want to know the value of gamma I want to find out the value of V by C. So let me square it and take inverse of it. If I square and take inverse of it I will get 1 upon gamma square square it so this under root goes away take inverse so 1 minus V square upon C square comes on the numerator this becomes 1 minus V square by C square. So I can solve this particular equation so this becomes V square by C square is equal to 1 minus 1 by gamma square which means V by C square is what we call as beta square in our notation so I can write this as beta square as I am sorry there is no unit here beta square is equal to gamma square minus 1 upon gamma square. So this is the expression which I have written in this particular transparency that beta is equal to under root gamma square minus 1 upon gamma square. So this expression gives me the value of beta which is V by C using the value of gamma that we find. In the present case gamma is 3 so I substitute it here this becomes 3 square minus 1 so this becomes under root 8 divided by gamma square which is 9 this 9 can be taken out of the under root sign so this becomes under root 8 by 3. So this beta is under root 8 by 3 which is V by C. Now once I know the speed I can always find out what will be the momentum of the particle and the momentum of the particle as we know is given by the momentum is equal to gamma m naught V. We know gamma is equal to 3 m naught is m naught and V is under root 8 by 3 C we have just now found out. So I have substituted this in this particular expression so P l l is again because I am describing this in the laboratory frame is equal to 3 multiplied by m naught multiplied by a root 8 by 3 C which gives me root 8 m naught C. Of course as I said I could have found out this particular expression also by using directly the energy momentum relationship which is E square is equal to P square C square plus m naught square C to the power 4 this is very commonly used relationship. So from this particular thing I have to evaluate momentum I know the energy so P square C square will turn out to be equal to E square minus I take this on the other side minus m naught square C to the power 4. E being 3 m naught C square E square will become 9 m naught square C to the power 4 so I substitute it here I have to subtract this expression minus m naught square C to the power 4 so this becomes 8 m naught square C to the power 4 C square I will cancel out and P I will get under root m naught C. So what we have done so far I know the energy the initial energy of the system which happens to be just one particular particle which now eventually decays into two particles. So this particular particle has a total energy of 3 m naught C square and it has a total momentum of under root 8 m naught C. So once this particular particle decays into two particles the total energy must remain same similarly the total momentum must remain same because this is purely a one dimensional problem so I have not taken the x axis and y axis I have not described it in the vector form all I need that this particular value of momentum must be concerned. So I will be able to write the following equations which I have written in the next transparency. Once we have found out what is the energy and the momentum of the original particle what we have to do is to write the energy and momentum conservation equations. We have seen in fact it has been given in the problem that the initial energy is of is equal to 3 m naught C square there is only one particular particle so there is only one particular particle system. Once there are two particles which have resulted out of the decay of this particular particle if even l and e2l are their energies then 3 m naught C square must be equal to e1l plus e2l. Now corresponding to 3 m naught C square I have seen that the value of the momentum is under root 8 m naught C remember this is the original particle so I have used capital m naught here. This must be equal to the sum of the momentum of the two particles we have been told that the particles move along the same direction as the original direction so I can write just in a scalar form I do not have to use vector form because as we have said this is purely a one dimensional problem. So this must be equal to p1l plus p2l where p1l is the momentum of the first particle p2l is the momentum of the second particle as I have said e1l is the energy of the first particle e2l is the energy of the second particle. In a traditional classical mechanics this would have been a very very simple trivial problem to solve all you have to do is to express this energy into momentum in the form of momentum or this momentum into the form of energy and just solve these equations you have two unknowns and two equations to solve it this would have been very simple mainly because in traditional classical mechanics in which there is no relativity the relationship between e and p are fairly simple of course in that case we know what we try to do is to conserve the kinetic energy that is not about that thing. Now in this particular case what we will find out that the energy and momentum relationships are little more involved which has been written in the next transparency. So the energy of the first particle is related to the momentum of the first particle by this particular expression e1l square is equal to p1l square c square plus m0 square c to the power 4 remember I am using small m0 because the energy and momentum of the particle that I am going to relate has a rest mass of small m0 this is not the original particle the original particle is not decayed the resultant particle which I have got which has a rest mass of m0 and fortunately in this particular problem the two particles have the same rest mass m0 that is what has been given. So I am using small m0 here in both these expressions exactly similar equation I write for as a relationship between energy and momentum of the second particle e2l square is equal to p2l square c square plus m0 square c to the power 4 remind that this l has been written just to make sure that I am talking of laboratory flame of reference. As we will see that is not very easy to solve these equations because if I try to express this particular equation in that particular form let us say let us try to express in energy in terms of the momentum then I have to take under root of this particular expression to find out what is e1l so I must write e1l is equal to under root of p1l square c square plus m0 square c to the power 4. Similarly for e2l I have to write under root p2l square c square plus m0 square c to the power 4. When I write in the conservation of energy expression I have to write 3 m0 c square is equal to this plus this so this becomes this under root plus this under root. If you can solve this equation of course you will get energy and momentum and in fact we will get p1l and p2l taking the other to other equation also into consideration writing p1l also in terms of the the initial momentum I will be able to solve it but as you can see because of the presence of these square roots etc these are not very straightforward equations to solve. On the other hand this equation becomes fairly this problem becomes fairly simple if I go to the center of mass of frame of reference. So now let us attempt to solve this particular problem in the center of mass frame of reference. Of course here the problem is fairly simple because there is only originally one particle. So if I have to find out what is the velocity of the center of mass it is going to be the velocity of the particle the initial particle that I am talking about. So I go to a frame of reference in which the initial particle is at rest and because in this frame the particle is at rest the momentum is 0 there is only one particle so that automatically becomes the C frame or the center of mass frame of reference. So let us describe let us go to a frame of reference in which the original incident particle or undecade particle not the incident but undecade particle was at rest. So this is what I have said let us go to C frame as there is only one single particle we shall have the following conditions I know very clearly because if I have gone to the frame of reference of that particle itself the particle is not moving in that frame therefore its total energy has to be only the rest mass energy there cannot be any other energy. Therefore the total energy in C frame will have to be just m not C square and of course if the speed of the particle is 0 then momentum is also 0. So therefore in the C frame the initial value of momentum is 0 which has to be if it is to be called a C frame and the total initial energy is just capital M not C square. Now let us try to see if I have to conserve energy and momentum in this particular frame of reference remember because the rest mass of the two particles that I am getting as a result of decay happened to be same the only way these particular particles can decay is when they move back to back it means one particle goes in this particular fashion another particle goes in this fashion opposite fashion that is the only way that they can conserve and make momentum 0 because remember initially the momentum is 0 so the final momentum also has to be 0 if the both particles move in the same direction there cannot be momentum cannot be 0 we have been told that they move only in one line so there is only possibility is that one goes in this particular fashion another goes in this particular fashion and because their rest masses happen to be same therefore their energy has to be equally shared because in order to conserve momentum both must have same momentum obviously and energy and momentum relationship depends on their rest masses their rest masses being same in forces that their energies are also same let me just explain this particular part little bit more so we have e square is equal to p square c square plus m naught square c to the power 4 if first particle p is same as the second particle m naught is same as the second particle then e of that particle also has to be same as the second particle therefore the two particles in the center mass frame of reference must have the same energy as well as the same momentum same momentum because the initial momentum was 0 so final momentum also has to be 0 and energy have to be same because their rest masses also have to be same therefore we can see that the problem is extremely simple it means each one of the particles must be having the energy of half m naught c square so this is what I am written here that m naught c square should be equal to 2 this is I can write as gamma c m naught c square so this m naught c square these the total energies I have to be equally shared and this energy of each particle can always be written as we have just now said gamma m naught c square and because the center of mass frame of reference so I am writing gamma c so each particle must be having the same energy so this 2 must be equal to the total initial energy in this frame which is m naught c square so this is what I have written m naught c square is equal to 2 gamma c m small m naught c square it immediately gives me the value of gamma c which is c square cancels here m naught divided by 2 small m naught once we have found out the value of gamma c I can immediately find out what will be the speeds of this particular particle in this particular frame of reference by using the expression which we have just now written earlier so the two particles will move in opposite direction with the following speed this expression is the same expression which we just now have worked out a little bit earlier and if I substitute the value of gamma c which I have obtained I will get that the speeds of the particle will be like this by will be given by this expression one moving in plus direction another moving in minus direction you can call it x direction one moving in plus x direction another moving in minus x direction so these are the two velocities but what we have found out are the velocities in c frame of reference if I have to find out 11 laboratory frame of reference all I have to do is to do a simple velocity transformation I have to know what is the velocity of the c frame which by the way we have found out earlier once we know the velocity of that particular frame take care of proper science and you can transform back and obtain the values of speeds or the velocities in laboratory frame of reference so we transform it to c frame then brought it back to L frame but remember the problem is very simple now we don't have to deal with all those under roots and trying to work out and trying to solve those equations the problem is essentially very simple so these I have written these speeds have only to be transformed back to L frame taking care of appropriate sign the speed of c frame in L frame needed need to apply this transformation this was already found out earlier if you remember earlier case which we have done here it already found out that beta is equal to under root 8 by 3 so from that we can find out v is equal to under root 8 by 3 c once we know the relative velocity between the frames we know the velocities of the particles in c frame use a velocity transformation bring it back to level triple will not work it or further I think this one can do it simply now let us go to a slightly more involved problem it's not really involved it's slightly more in the sense that instead of one initial particle we have two particles and things are not really in one dimension things are form to two dimension so let us read the problem an electron of total energy 1.4 MeV again when you say total energy it means it includes the rest mass energy this total energy is 1.4 MeV m is 10 to power 6 electron volt collides with another electron which is at rest in L frame so in laboratory frame of reference there is one electron which is at rest another part of electron comes and hits it collides it or you can get scattered whatever you want to call it after the collision the target electron it means the electron which is originally at rest is found to get scattered at an angle of 45 degrees but not in the laboratory frame but in the c frame in the center of mass frame of reference so the problem involves both laboratory frame and c frame so in that sense the problem is little more involved so after the collision the target electron which was originally the electron which was at rest is found to get scattered at an angle of 45 degree c frame so from the nature of the problem itself wording of the problem itself it's clear that we have to be talking about c frame find the energy and momentum components of the target electron after the scatter in s and in c and L frame so after the scatter after the scattering has taken place what are the momentum and energies of the electron in in the frames of course rest mass energy of the electron has been given as 0.51 so I think the problem is clear that this one particular electron which is being hit by another electron one electron is at rest another electron comes and hits here it hits it the electron which was at rest is found to move at an angle of 45 degrees in center mass frame of reference of course because when you are talking of 45 degrees it means it's no longer a one dimensional problem it has to be you have to work out in two dimension and then you have to find out what will be the energy and momentum of the particle probably you can work out for both the particles if necessary or target electron in laboratory frame of reference as well as c frame of reference what are the issues in this thing first of all it's a two particle system from the beginning as we have said that's no longer a one particle system so like in one particle it was very easy to find out what is the center mass frame of reference because you have to just go to the reference frame of reference of that particle itself in which this particle happens to be at rest so here it's not so obvious but we have to work it out the first method which is slightly I mean it's straight forward method but little more longer method in fact we will give you a comparatively more straight forward method is let's assume that c frame travels in the laboratory frame of reference with a speed v then find the momentum and energy of both the particles in the c frame so assume go to an arbitrary frame of reference which moves relative to L frame with a speed v then find the momentum and energy of both the particles in the c frame then take the sum of the momenta put it equal to 0 because I am looking for that particular frame of reference in which the sum of momenta is 0 so once I put I take momentum of first particle and moment of the second particle in a frame of reference which is moving with a speed v relative to L and equate this particular momenta sum of the momenta to 0 and solve for v that will be the velocity of the center mass frame so this is very straight forward standard method of finding out even if you had n particles that's what I will do in principle if I want to do in a most simple fashion that find out the moment go to a particular frame of reference which moves with a speed v transform all the momenta to that particular frame of reference takes summation of the momenta put it equal to 0 that velocity that will obtain will be the velocity of this center mass frame of reference so let us first work it out that that way before we go to little more involved little more trickier way but simpler method so let us first find out the total initial energy of the electrons as we have seen that one electron is at rest in the laboratory frame so I am doing this particular problem first in L frame so that energy is 0.51 electron mev and the other electron which is at which is coming and hitting it that has a total energy of 1.4 mev so total energy in the laboratory frame is 1.91 mev so this is the total energy which is available to me in the laboratory frame of let us work out momentum second particle is at rest and the target particle is at rest in the laboratory frame of reference so obviously its momentum is 0 now we have to work out the momentum of the other particle which is coming and hitting it here find out what is the momentum of that particular particle I know the energy of that particle which is 1.4 mev I use the same standard expression e square is equal to p square c square plus m naught square c to the power 4 m naught square c to the power 4 will be just 0.51 square 0.51 mev square so this is what I have written 0.51 mev square so this was p square c square plus m naught square c to the power 4 and this was e square which is the total energy so I am solving for momentum like we did for the earlier problem so this becomes 1.4 square minus 0.51 square only thing which I have done I have taken a special units here so that this particular normally it should have been p l c but I have taken units of mev by c so this value of c I need not write here so this is very conventional unit our traditional unit sorry in which we express momentum in the units of mev by c energy by c energy we express in mev or gev or whatever depending on the type of problem and express momentum in mev by c or gv by c so the momentum of this particular particle is one point if you just work it out turns out to be 1.304 mev by c of course the momentum of the second electron is 0 that is what we have just now said now transform it go to a frame of reference which is moving with a speed v in the L frame I use momentum transformation momentum transformation is standard gamma what is the original momentum px p1 x prime is equal to gamma px minus v e by c square this is the momentum transformation equation momentum we have just now find out 1.304 I am writing the unit to make it clear minus v is the speed relative speed between the frames the energy is 1.4 mev divided by c square for the second particle the initial momentum is 0 so gamma multiplied by c into 0 minus v into its energy is only its rest mass energy so v multiplied by 0.51 mev divided by c square in fact we just take the sum of these two put it equal to 0 solve for v that will give me immediately the velocity of the center of mass frame of reference that is what I have written for c frame p1 x prime plus p2 x prime is equal to 0 so just sum these two things and work out for v you will get v is equal to 0.683 c from this particular way of working it out and then once you know the velocity you can find out gamma once you have found out gamma then you can transform the energy and find out what is the energy of these particles in the center of mass frame of reference then center of mass frame of reference initials some of momentum is anyway 0 new energy momentum but as I told there is a slightly quicker method to work it out by using the convention of the four vectors so let us now adopt method 2 of working out this particular problem going to center of mass frame of reference by using the concept of four vectors as we will be seeing that though this is slightly tricky but it is much more simple so this is what I said method 2 one can use the fact that the length of four vector will be same in all the frames we have discussed earlier when we described the concept of four vectors that the length of four vectors is what we call as a four scalar once we change the frame of reference the individual components of the four vector may change but its length is unchanged that is a four scalar so once I go from one frame one inertial frame to another initial frame of reference length is going to be same so what I can do now calculate the length of energy momentum four vector in the laboratory frame of reference for the system of the particles and I know that this particular length is going to be same in the center of mass frame of reference and I know in the center of mass frame of reference summation of momentum will be 0 so it will have essentially only the fourth term which is energy if you remember the four terms of the energy moment of four vectors where p1 p2 p3 and or rather px p by pz and fourth term was i e upon c so fourth term depends on energy let us just work it out and see how simple it this particular problem becomes so first let us write the length of the four vector for this summation for the system of the particle so first three terms you are taking the length so it will even square plus a2 square plus a3 square plus a4 square first three terms are only the momentum terms so it essentially means the first three terms will give me a summation of summation of pli square where l is the momentum in the laboratory frame of reference some over all the particles in this case there are only two particles so what I will do I will add the momentum of the two particles and this particular term will be obtained from that similarly this particular term depends on the summation of the energy of the two particles because I am applying this particular I am calculating the length of the energy momentum four vector for the system of the particle for two particles together not for the individual particles so when I am writing for the two particles together this summation will be the sum of the energies of the two particles first particle has energy of 1.4 MeV the second particle is at rest it has only Rasmus energy which is 0.51 MeV so summation of ELI will be only 1.91 some of first energy of first particle plus sum of energy sum of energy of first particle and the second particle first particle having energy 1.4 second particle having energy 0.51 similarly here summation of the momentum magnitude of the momentum for the first particle which is 1.4 and for other particle is 0 so this becomes one I am sorry the moment of the first part is 1.304 for the second particle is 0 so it just becomes 1.304 square I calculate this number this turns out to be an imaginary number the length of four vector can always be imaginary we have discussed these problems earlier this will be given by 1.396 I MeV by C now I know if I go to any other frame of reference this length for the same system of particles is going to remain same but if I go to a specific frame which is called center mass frame of reference in that summation of momentum will be 0 it means this particular term would be 0 so if I want to write in center mass frame of reference this term is 0 and only the energy terms will be present okay so it is much easier to find out the energy in the center mass frame of reference let us see how first let us realize that because the two particles that I am talking are only electrons so what will happen in the center mass frame of reference one particle comes like this another particle comes like this both are electrons they have to come in opposite direction because they have to make the total momentum 0 so their momentum must be in opposite direction now they get scattered one of this electron goes this way another electron goes this way we have been told that this angle is 45 degree this has been given in the problem that this angle is 45 degrees all right now the only way because initial momentum is same as the final momentum so if this electron goes this way making an angle of 45 degree the other electron has to go opposite back to back just like in the earlier case except for the fact that it's not in the same line but now treated at an angle of 45 degree but otherwise this electron has also to go back into back to back because the rest masses of the two electrons are same it's an electron so rest mass is same okay therefore its energy has to be same and the moment also has to be same so once we realize that this is what is going to happen then I realize that this particular energy must be same as this particular energy as we have discussed earlier now what will be the summation of the what will be the length of the four vector in C frame summation of Pi 0 so it means it must be minus summation of ECi square by C square I have put the symbol C here to demonstrate that this is in the center of mass frame of reference so this particular thing because the PC is zero some of the I'm applying again I'm reminding I'm applying for the system of the particle and for the system system of the particle in center of mass frame of reference summation is zero so it contains only one term which is this and this must have exactly the same value as we have done earlier because the length is going to be same in this frame for the same system of particles so I just evaluate this particular thing and in principle I will get summation of ECi and I realize if I take under root that this is going to be the sum of the two energies and because energies are going to be shared equally by the two electrons so each one of them will have a energy of 1.396 divided by 2 let me repeat this length of the four vector has only one term which is here which contains the summation of the energy of the two electrons the summation of the energy of the two electron and with energy being equal for both the electrons it means ECi is just this particular summation is just the double of the energy of the individual electron so this 1.396 if I divide by 2 I will get the energy of the two electrons this is what I have written the energy of the two scattered electrons and the magnitude of the moment are same we therefore get ECi is just as 1.396 divided by 2 is equal to 0.698 MeV so each of the electron will have this much energy because of the symmetry involved in the center of mass frame of reference the two particles have to go back to back and both being the same particle the same same rest mass have to have the same energy and of course they have to have the same momentum I can calculate what will be the momentum of the particle by just using exactly the same relationship and I will get that the momentum of the individual particles individual electrons will be 0.477 MeV by C in C frame of reference which is center of mass frame of reference I have found out energy and momentum now in C frame laboratory frame of reference make a transformation again you have to find out the speed of C frame with respect to L frame of course if you have adopted the second method we have so far not obtained the speed of the first method we found out the speed of the center mass frame of reference but in this particular method we have not yet found out but we can now find out because in this particular C frame the energy of the electron is 0.698 MeV while its rest mass energy was 0.51 MeV so the gamma will be 0.698 divided by 0.51 once I know gamma I can calculate what will be the speed is what I have done of course before that you know let me also tell that we have to calculate the x component and the y component of the momentum because this now making angle of 45 degrees and because it's an angle of 45 degrees so x component and y component will be just whatever is the momentum divided by root cos 45 sin 45 both are 1 upon the root so these are the x component and the y component of the momentum and I must you know we should be clear that we have to transform not just one momentum we have to transform x component of momentum y component of momentum and also the energy of the particle to L frame so as I said relative speed of C can be obtained by 0.698 divided by 0.51 which gives me the value of gamma is equal to 1.369 and I can calculate the beta as we have done using the expression gamma square minus 1 divided by gamma square which gives me the same value which I have obtained by the first method which is 0.683C I know the velocity I know px I know py I know e in center mass frame of reference apply transformation equations go back to laboratory frame of reference you will know all the energy all the momentum laboratory frame of reference if you are not very sure you have done everything correctly again apply conservation of energy and momentum in that frame they have to be obeyed in that frame of reference if you have not made any mistake so that's what I have said the energy and momentum can be obtained in laboratory frame using energy momentum transformation standard transformation we have been using those equations earlier one can also ascertain that the energy and momentum are also conserved in the L frame which they have to that's not a part of the problem to show that but if one wants to be very sure to see that I have done no mistake this is the way one can quit I just wanted to point you out one particular thing that once we did when described experiment in center mass frame of reference the energy that we got was much less in the laboratory frame of reference it was 1.4 plus 0.51 MeV while in the case of the center mass frame of reference is 1.396 MeV so just illustrates this particular fact that the if we have to design this particular experiment then in the center mass frame of reference the energy that is actually being used is only 1.396 MeV while in the laboratory frame of reference it's 1.91 MeV this is what sort of illustrates what I have been telling earlier which I told earlier that many times it's simple it's energetically favorable to design the experiment in center mass frame of reference so you do not spend energy in accelerating or I am not letting this center mass. Now let's take one more problem in which also we are using the concept of four vectors and the concept of center mass frame of reference it's also very simple problem but involves many more particles. So problem is as follows that there is a proton P which has a kinetic energy K this problem has been given in terms of kinetic energy most of the time the problems are given in terms of the total energy in this it happens to be given in terms of kinetic energy no issues we know how to convert it no problem we also know the expression between energy kinetic energy and the momentum so we can use that expression. So we have a proton which has a kinetic energy K and this is incident on another proton which is rest in the laboratory frame of reference even if I would not have said L frame would have realized that this is not a center mass frame of reference because one particle is at rest another particle is moving so obviously the center of mass cannot be at rest so this is obviously not a center mass frame of reference. Now this proton when it's incident on the other another proton results into four different particles after the interaction four particles are found not two particles unlike the earlier case the four particles which are found three of which them are proton and the fourth one is an anti proton just to ensure anti proton has the same mass as a proton same rest mass as a proton but it has a charge which is negative to it has a charge which is negative protons are positively charged particle while anti proton is a negatively charged particle you get three protons and one anti proton. Now question is little more tricky what should be the least value of kinetic energy K what should be the least kinetic energy that you must supply to that incident proton so that this particular reaction becomes possible see from two protons you are generating four particles three of which are protons and one of them is anti proton so obviously if both the protons were at rest you could not have done it because the energy that was available to you was only two m0c square where m0 is the rest mass of the proton and you are getting four particles each one of which has the rest mass of m0s rest mass of proton it was not possible obviously you require certain amount of energy now you cannot just apply conservation of energy because the momentum also has to be concerned but if you go back to the center of mass frame of reference there is certain amount of easiness because in center of mass frame of reference I am very clear initial momentum is 0 and in that particular frame of reference to require least amount of energy is that I can create all the four protons at rest if I think in terms of center of mass frame of reference I have two protons which are coming back to back like this and hitting it they have to have equal momentum because this is in mass frame of reference they have the same rest mass now if you are giving them too much of energy you have four protons which may move anywhere else four protons have three proton plus one anti proton four particles but if I want this particular thing to have the least amount of energy I can have such a velocity such that these four particles are all at rest and because all the four particles are at rest then momentum is still 0 this is what I am trying to explain I have written the next transparency the least energy required would be when all the four particles are created at rest but that's only possible in the C frame of reference because it's only in that particular frame of reference that the momentum is 0 in laboratory frame the two particles the four particles cannot be created at rest because then momentum will not be conserved because initial momentum is non-zero it's only in C frame that the initial momentum is 0 so you can imagine that all the four particles are created at rest so we go to center mass frame of reference the energy is C frame corresponding to the case when all four of them are at rest must be equal to four M0 C square where M0 is the rest mass of the proton because each four all the four particles have the same rest mass M0 and they are not moving because I want the least amount of energy so the total energy which will be generated in the center of mass frame of reference has to be four M0 C square therefore the square of the length of the four vector in this frame because momentum 0 is minus E square by C square so it must be minus 16 M0 square C square this will be the length of the four vector in the center of mass frame of reference after the reaction is over when we have got four particles what will be the length in the laboratory frame of reference before the interaction we have momentum which is not equal to 0 I can find out what is the momentum there is only one particle which is moving so I can find out what will be the total momentum of that particular particle knowing its energy I will know the energy of the first particle I know the energy of the second particle I can find out the length of the four vector just like I did in the previous problem because there is only one term so I have not written summation here here of course I have to write the totally I have not put the summation sign makes things simple so this p square will be given by K square by C square plus 2 M0 K this is a standard expression relating momentum to the kinetic energy this is what I have used earlier the total energy energy of the first particle which is coming with the speed is kinetic energy plus rest mass energy so K plus M0 C square there is another proton which is at rest which has the energy of just M0 C square so total energy is K plus 2 M0 C square this is summation of E I have not written summation sign here this is the total length of the four vector so that length of the four vector in laboratory frame of reference is minus 2 M0 K minus 4 M0 square C square now this length of the four vector after collision has to be or after the interaction has to be same in the length in the laboratory frame of reference and has also be the same in the in the center mass frame of reference so what I do I take the length of the four vector of the system of the particles before interaction in L frame and equate it to the length of the four vector after reaction in C frame so this is what we have been I am taking before interaction the length in laboratory frame and equating it to the length of the four vector in after the reaction in C frame which I know is minus 16 M0 square C to power 4 if I want the least energy this is what I have written equating the length in L before interaction to the length in C frame after interaction I must have minus 16 M0 square C to power M0 square C square given by this particular expression I solve this K turns out to be equal to 6 M0 C square so this is the least K needed in laboratory frame so in laboratory frame of reference this particular particle must have a kinetic energy of at least 6 M0 C square in order that reaction becomes possible. In summary we have discussed some problems in center mass frame of reference and demonstrated how it becomes easy to solve these problems and we also gave some examples of the use of four vector concept thank you.