 Okay, good morning let us start, I will start with answering few questions which we have received through our forum. There is one question which probably is not very well framed but you know let me just I think I understood the question which is what will prevent tunneling increase in height or increase in width of the potential probably what it means that you know how the tunneling probability will decrease whether by increasing height or by increasing the width in fact it is done by both I had given you the expression of actually inverse of transmission coefficient you can see both width as well as the height of the potential barrier comes there so it depends actually on both. The next question is that can a particle with zero rest must travel more than the speed of light no because as we just now I mean last time saw that particle I mean of course we have not introduced photon which I will do today will travel exactly with the speed of light there is no other way I mean I will show that this is in order to maintain the consistency in the equations of special theory of relativity it is necessary. Then there is another question is does a Michelson-Modell experiment uphold mass energy equivalence see mass energy equivalence is something different as today will I will be discussing okay. So as far as Michelson-Modell experiment was concerned it was only to test whether the speed of light is a frame dependent quantity so I at least I cannot see any direct relevance with the mass energy equivalence see mass energy equivalence comes when we actually take the postulates of special theory of relativity and in order to maintain those postulates we require some change in some of the basic equations and then mass energy relationship comes in. Then earth is a inertial or non-inertial frame of reference that I think we have discussed number of time that is actually non-inertial frame of reference I mean there is nothing like real inertial frame of reference but for many practical purposes as we can treat this as inertial frame of reference. Then you know there is about length contraction whether it has a physical significance or not of course I have not talked about length contraction but somebody asked about time dilation I sort of introduced time dilation at that time I have mentioned briefly about the length contraction force length contraction is always real if you are going from one frame to another frame of reference so that is absolutely no issue see what you are seeing as the proper length of a particular rod in a frame of reference and that length has to be along the direction of the relative velocity then you will always find that whenever a measurement is done to measure the length of that particular rod in a different frame of reference you will find it contracted of course this contraction will be visible only when you know the relative velocity between the frames is very close to the velocity of light when I had said that even if it is 0.1 C which is fairly large velocity there also you will not be able to I mean you are making only 1% correction in the classical equations. Then I think there are two more questions why the Galilean transformation failed on the basis of electromagnetic that is what I had explained that because according to Galilean transformation we will have the classical relative velocity formula and application of that relative formula will tell you that the speed should be different in different frames which as I said has a problem because first of all speed of light is given in terms of basic fundamental constants but more importantly then you know there is a Michelson-Moldy experiment which shows that the light does not seem to be frame dependent. Then is it possible to show a video clip unfortunately I do not have any video clip of Michelson-Moldy experiment arrangement I mean I would say is the best thing would be that you know you look and do a Google search there is a possibility that you may see some of the clips. I think these were some of the questions so now let us come back to my last lecture in this particular thing it was pleasure discussing and talking to you. So this is what we discussed last time we discussed that time is frame dependent that is what Einstein thought that is probably the only way to maintain that speed of light is frame independent is to make time frame dependent. So was one thing which we discussed last time in detail then we discussed the actual orange transformation which replaces the classical Galilean transformation then from that we obtained the velocity transformation which ensures that the speed of light is same in all the frames and then we say that this is the most general velocity transformation. Of course again like Lawrence transformation you can see that only when the relative velocities are of the order of speed of light then you will start seeing its effect or its difference from the classical formula otherwise the classical formula of relative velocity if the velocities are not that large would still hold good. Then I worked out one or two examples rather one example to see how we calculate a relative velocity and then in answer of one of the questions I mentioned that you know if I take one particular particle which let us say in Earth's frame of reference is driving the speed point 6C or 0.8C and there is another particle which is coming with particle with a speed velocity of light. Then in that frame of reference also when we transform that particular person will find the speed of the particle to be speed of light. So this is what was actually required when we are talking about the velocity transformation because that was the postulate. Now we go to one step forward we have only tried to ensure the second postulate of special theory relativity which says that speed of light is same in all the inertial frames of reference. We have not tested the first hypothesis or first postulate which says that laws of physics are same in all the inertial frames. Now question is that is this enough is just changing Galilean transformation to Lorentz transformation is enough and is that the only modification that we need. The answer to that question is that no that is not enough. We require some more changes so that my laws of physics became consistent with the or rather my expression I will not say laws of physics but my expressions turned out to be consistent with the first postulate which says that laws of physics should always be frame independent. So to explain this thing I will take one very simple example this is often taken most of the textbooks it is taken. This is about the law of conservation of momentum which we believe that is a fundamental of physics. So if we have only Lorentz transformation and the velocity transformation as derived from the Lorentz transformation then if we take a normally standard classical in elastic relation we will find that in this particular case if the momentum is conserved in one frame it may not be conserved in any other frame of reference. It means the law of conservation momentum will turn out to be frame dependent which I cannot allow because of my first postulate which says that laws of physics should remain same in all the inertial frame of reference. Then essentially means that we have to change the definition of momentum and eventually we will find out that we will have to change the definition of energy then only you will maintain the universality of law of conservation of momentum and law of conservation of energy it means in all the inertial frames momentum and energy will turn out to be conserved. If they are conserved in one frame in another frame it will also turn out to be conserved. So let us take this very simple example and let us try to see that just velocity transformation derived on the basis of Lorentz transformation is not good enough. So this is what I have written in this particular slide need to redefine momentum. So essentially this is to say that you have to relook at the definition of momentum. I said this is a completely inelastic collision which we probably know right from classical theory in high school people know this particular problem very very well. This is in which let us suppose this is a frame S in which we are observing this particular collision. Of course I have taken the speeds which are very large so that you know we can see the relativistic effects. So let us suppose there is one particular mass which is travelled to the right which is travelling to the right hand side with a speed of 0.6 C and there is another mass which is travelling to the left hand side with a speed of 0.6 C and the two masses are same. Then they collide and they get stuck to each other and they come to rest. As I said this is called a completely inelastic collision in the traditional classical mechanics. In this case we always tell the students that momentum of course will be conserved but the energy will not be conserved because this particular particle will have energy classically speaking half MV square and this is half MV square. So overall the total energy will be equal to MV square but after they collide they come to rest and when they come to rest the kinetic energy turns out to be 0. So this energy is not conserved. On the other hand momentum turns out to be conserved because momentum turns out to be m times V and this is also m into V and remember momentum is a vector quantity. So this momentum is in the negative x direction, this is in the plus x direction. So overall momentum of the system will be 0 and when they have come to rest then of course the velocity is 0 so momentum is 0. So momentum is conserved and energy is not conserved. Of course in that case also we had we tell them, we tell the students see remember when we are talking that energy is not conserved we are talking only of the mechanical energy it is only the kinetic energy which is not conserved. Actually the energy is converted into a different form which is not really the mechanical in nature. For example when they collide then some sound may come out or the material may get heated up somewhere this particular energy has to be utilized. But these are not really in the form of mechanical energy overall energy we do expect to be conserved. But in the traditional classical mechanics the mechanical energy is not conserved in this process. Now let us apply my relative velocity formula of course as I have just now said that energy and is not conserved classical energy is not conserved momentum is conserved. Now let me observe this particular collision from a different frame and let us assume that this particular frame is the one which is also moving to the right hand side with a speed of 0.6C. So let us suppose this frame in which I have described this collision is a frame S and let us now go to a frame S which is moving in this particular direction plus x direction with a speed of 0.6C. We will use the relative velocity formula which we have just now derived or rather derived last time and see that in this particular frame of reference momentum will not turn out to be conserved. So as far as this particular S frame is concerned this momentum turns out to be conserved. Let us look into this thing. So in S frame as we have agreed that initial momentum is 0 when I say sum of initial momentum of the two particles as I just now said because one moves with the same velocity on the left hand side another moves to with the right hand side and momentum being vector if I take the sum of the momenta of the two particles that will turn out to be 0 and final momentum is also 0. So momentum is conserved. At the moment I am looking only at the conservation of momentum aspect. Now let us view this collision from another frame S with a velocity v is equal to 0.6C using the velocity transformation. Yesterday I worked out a problem how to find out the relative velocity how to change the velocity to a different frame of reference. Do the same thing now for this particular problem. So I will try to find out what will be the velocity of this particular particle in this S frame. I will also try to find out what will be the velocity of this particular particle in this S frame. Then what is the velocity of this particular particle which is now at rest in S frame in S frame and then try to apply conservation of momentum to see whether momentum is conserved or not. So let us go to S frame of reference. Now we start looking at the velocity of the first particle. Let us look here because I am going to frame of reference where v is equal to 0.6C this is what I have said. So it is 0.6C is the velocity and this particular particle which was moving on the right hand side also has a velocity of 0.6C. So it means ux is also equal to 0.6C. If I want to find out the velocity in this particular frame of reference which is travelling with the speed of 0.6C in frame S then ux will be equal to ux minus v divided by 1 minus ux into v upon C square. ux is equal to 0.6C v is equal to 0.6C denominator will become 0. So ux prime would be equal to 0 which is understandable because that particular frame is travelling exactly with the same speed of the as the particle. So there is nothing surprising to see that the velocity of this particular particle turns out to be 0. So this is what I have written. This is prime of the first particle I have written x because of course there is only x component here. So I need not have written but you know nevertheless I have written it. So this is 0.6C minus 0.6C I have substituted here but that is not necessary it becomes 0 this is as expected. Now exactly same thing I will do for another particle second particle thing is that there instead of 0.6C it will be minus 0.6C. So numerator will become minus 1.2C. So this is what is written in the next transparency minus 0.6C minus 0.6C and here of course it will become 1 plus because ux is negative. So you have 1 minus ux into v and ux be negative that becomes 1 plus ux into v upon C square. So denominator will become 1.6C 36 and I had exactly worked out this type of problem yesterday. So second particle speed in this S frame of difference will turn out to be minus 1.2 divided by 1.36C. Now let us look at that particular particle which is at rest. For that particular particle you have this 0 here and because you have 1 minus ux into v upon C square ux being 0 so denominator this particular term will not be present. So after the velocity of the combined particle after the collision ux turns out to be 0 because in S frame that particular particle is at rest minus v divided by 1 minus this. So this is also equal to minus 1.6C this is also consistent with the standard classical result because remember this particular denominator term this particular term turns out to be 0 and numerator was the classical term. So now according to S prime observer first particle is at rest. Second particle travels with this speed third particle which is a resultant particle after the collision will travel with the speed of minus 0.6C. So I apply conservation of momentum it means first particle has of course 0 momentum in S prime because its speed is 0 second particle will have moment of minus 1.2 divided by 1.36 times C multiplied by mass. Here it will be 0.16 and of course the mass of this particular particle will be 2m so minus 1.6 into 2mC. As you can see here this will be 1.2 while here it will be 1.2 divided by 1.36 it means the momentum initial momentum will be different from the final momentum. So according to S frame of reference when this collision occurred momentum was not conserved if he applied only the velocity transformation equation. But this is an outcome which I cannot accept because this violates the first postulate of relativity which says that all laws of physics should be same in all inertial frame of reference. It means just applying relative velocity formula to this particular problem was not correct we need something more basically we need to redefine momentum. The initial and the final momentum in S prime is given by this particular thing. So initial momentum is minus 1.6 and of course the final momentum is minus 1.2mC the initial momentum will be minus 1.2 because momentum of the first particle is 0 so this will be the initial momentum this is the final momentum which is not conserved according to S prime. Now unfortunately I do not have time to give you little bit more logic on how momentum is redefined I will just give you the expression of momentum and we will emphasize more on how to use this particular expression of momentum and eventually energy to solve certain problems. If at one can give some certain amount of logic to come to this particular why momentum has to be defined and how the universality of conservation of momentum and conservation of energy is maintained with this new definition. This is something which I will not be able to touch here and again I say that you know if you are more interested you know can I have also a series of video on YouTube you can probably watch or you can look at any other textbook this is something which is covered in most of the textbooks. So now we go to a new definition of momentum. First of all we define a quantity which we call as a rest mass it is a mass which is measured in a frame of reference in which the particle is at rest. So we define a quantity which is called m naught then we define momentum as m naught times gamma u. Let me just spend one minute to explain these gammas because there are so many gammas which are coming. See so long there is only one gamma. See earlier if you remember we defined gamma as 1 upon under root 1 minus v square by c square where v was the relative velocity between the two frames from between which we are seeking a transformation. Now I define two more gammas one what I call as gamma u and another what I call as gamma u prime. See remember when we are talking about the velocity transformation at that time we talked of three different speeds or three different velocities. First velocity was the relative velocity between the frames for which we reserved the symbol v then we said that a particle is being observed by the observers in the two different frames. When an observer in s frame observes this particular particle he or she notices its instantaneous velocity to be u. So u is reserved for the particle velocity and as I have been emphasizing earlier u need not be constant v has to be constant. Now the velocity of the same particle is being observed by another observer in an s frame of reference and what that observer finds the velocity of the particle will be represented by u prime. Again u prime need not be constant. Now if I use an expression exactly identical to this but instead of v I use either u or u prime I will call it gamma u or gamma u prime. Unfortunately we have to define these symbols to make equations look simpler otherwise you know the equations itself writing takes enormous amount of space. So here we are using u remember u and u prime need not be constant therefore gamma u and gamma u prime need not be constant as a function of time they would change. Well as far as gamma is concerned because gamma is dependent on v and v has to be constant gamma will be constant. So this is just to emphasize on the symbols that I am using. So suppose I know the speed of a particular particle in a given frame of reference this speed is defined as u then using this expression I can calculate what is gamma u. Of course if the speed of the particle changes gamma u will also change. But let us suppose we are talking at a given instant when the speed of the particle is u then I calculate gamma u then I multiply this particular quantity called rest mass by gamma u multiplied by ux. Remember in the traditional definition of momentum we just have m times ux all that is getting multiplied is this particular factor gamma u. Now similarly as far as the y component of the momentum is concerned exactly the same thing we define this as m0 gamma u ui. Pz I define this as m0 gamma u z and because these are definitions so I have put three bars here to say that these are the way we define momentum in relativity. So say we define rest mass m0 which is the mass of the particle measured in a frame when the mass is at rest then we define px, py, pz using this particular thing. Now similarly if I have to find out the momentum in s frame of reference u will turn out to be different then I will use u prime this u will become u prime this will also become u prime this will also become u prime of course I will not be able to do again momentum transformation but as I said I will just discuss about the new definition of momentum and energy and try to use it to solve certain problems. So note in s frame of reference u will turn out to be different because the same particle velocity will be measured will be found out different of course that will be that can be found out by using velocity transformation. So then this u prime we will using this u prime I can calculate what is gamma u prime I use this particular gamma u prime multiplied by the x component of the velocity as observed in s frame of reference then m0 gamma u prime ux prime will give me the x component of the momentum. Similarly y component similarly the z component. Now the next question is that is this enough if we define this wave of momentum as I said I will not be able to discuss why we are going to this particular definition and how this particular definition ensures that momentum is conserved in all the frame of reference but it just let us say that we have introduced it let us say rather abruptly. Now question is that if we define this particular thing does that ensure that momentum will be conserved in all the frames. Unfortunately you can show that this does not happen just conservation of momentum will not work out we require further change. So I sort of ask only momentum conservation does the new definition of momentum guarantees universality of conservation of momentum the answer is no. In fact what can be shown that along with momentum you must conserve energy also then only universality of conservation of momentum will be maintained. So momentum and energy have to be conserved together in any process whether what I am calling this particular process as a completely inelastic in a process that was a process in which traditionally we thought energy is not conserved or rather mechanical energy was not conserved. So then Einstein proposed his well known formula of new definition of an energy which is essential to be conserved if momentum has to be conserved in all the inertial frames of reference. So this is what I am saying energy in relativity a totally new expression of energy is proposed in relativity which is E is equal to m naught times gamma u into c square this gamma u into m naught is often written as m and we often call it that mass is velocity dependent of course very large number of my colleagues who actually work in the area of particle physics which I do not do they always try to call when they see a mass they always call it rest mass and they say that is just m naught gamma u c square but you know that is I would not like to come into that particular thing. This is a very standard way of writing this particular thing that you write E is equal to m naught gamma u c square which is often written as m c square where m is m naught multiplied by gamma u. Now what is this energy? If I am having u is equal to 0 it means particle is at rest remember my gamma u was having this expression if this u is 0 gamma u is 1 also if u is very small in comparison to c gamma u is approximately equal to 1. So u is equal to 0 gamma u is equal to 1 it means momentum will be which is defined as this particular thing will become 0 because u anyway is 0. So this is consistent with the classical expression because in classically if a particle is at rest the momentum should be 0 and with the new definition of momentum also momentum turns out to be 0. But in this particular expression of energy if I put gamma u is equal to 1 actually I get E is equal to m naught c square unfortunately this does not match with the classical expression because classically we thought that the particle is at rest and let us suppose if there is no potential energy the total energy is 0. On the other hand with this new definition of energy even if the particle is at rest it will have energy which is called m naught c square I mean this is very very epoch making as I have discussed very very far reaching consequences this particular very very brilliant way of by Einstein to define energy. Similarly if u is very very small in comparison to c gamma u is also equal to 1 and p will turn out to be equal to m naught u which matches with the classical definition of momentum. So as far as the definition of momentum is concerned all that you have done is a multiplied by a factor of gamma u which is somewhat velocity dependent of course it will become significant only when u is very very close to c. So but this definitely yields to a classical definition of momentum in the classical limit but on the other hand for energy this does not have to turn out to be so in fact in the classical limit the energy will not be 0 but energy will be equal to m naught c square. This is a totally a new form of energy which does not have any classical analog. The new form of energy does not resemble any classically known form of energy it is a totally entirely a new concept of energy I mean that is why so much importance is attached to Einstein that about this particular thing. Now what Einstein proposed that any type of energy when you are talking whenever there is a energy gain or energy loss in respect to what type of energy that we are talking it will result into an increase or decrease of the mass. So for example if there is a particular particle which is at rest and I heat it I know that this particular energy has been supplied to this particular particle when the energy has been supplied to this particular particle even though this particular particle has is at rest in principle its rest mass would have gone up because it is having more energy. Similarly if a particular particle moves down it has emitted energy if it has emitted energy its rest mass should go down and if this particular particle starts traveling then it will gain some kinetic energy and that kinetic energy will be different as I will give you the expression from the new value of m c square and what was the original value of m naught c square which is the rest mass energy m naught c square we often call rest mass energy. Similarly if you are taking two particles let us say an electron and proton and they are at infinite infinite distance you bring them and you form a hydrogen atom and let us suppose this hydrogen atom is in the ground state then you will find the 13.6 electron volt energy will be released because this particular system of bound hydrogen atom in which electron and proton are bound has a lower energy then a system in which electron and protons are infinite distance away. So if you take a mass of electron and you take a mass of proton infinite distance away you make them hydrogen atom take the mass of the hydrogen atom the mass of the hydrogen atom will be smaller than the sum of the mass of electron and proton divided by c square it is a very very far-reaching thing. In fact as probably all of you know that many of the mass effect problems of nuclear physics where by measuring the mass of the nuclei we estimate what is the binding energy of nucleus now people have people attempt now with these things it is a totally a new concept of energy and we say because now mass can be expressed in the units of energy as these two are related through a fundamental constants. So generally I mean if you ask again a particle physicist particle physicist there is no difference between mass and energy because they are related by a fundamental constant. In fact to a particle physicist if you ask what is the mass of the electron you will say half Mev or 0.51 Mev Mev which is mega electron volt is the unit of energy but they are so used to expressing mass in terms of only energies what is the mass of the proton 938 Mev or normally they take approximately as 1 Mev. These are all units of energy but things that mass can always be converted into unit of energy because there is a fundamental constant c square which is involved. Now as I said we can define kinetic energy let us suppose there is a particular particle which has a it has a particular s mass energy and if this particular particle starts moving the difference between the two energy will be termed as the kinetic energy. So k is equal to mc square minus m0c square because remember this energy is always present when even when the particle is at rest. So whatever the additional amount of energy that this particular particle has obtained as a result of its motion that difference will be or that particular additional energy will be called kinetic energy. So this is what is the expression of kinetic energy but remember gamma u I can write as 1 upon under root 1 minus u square by c square and it should be minus m0c square. So this particular thing I can write as 1 minus u square by c square to power minus half then again you expand in the binomial expansion if you expand you will get this multiplied by plus half assuming that u is very smaller than c I neglect higher order terms. So this I can write as m0c square into 1 plus u square by 2c square remember this sign is plus because there is a negative sign here this sign is 2 here there this number is 2 here because there is a 2 here minus m0c square this m0c square will cancel it out and you will get half m0u square. So in the classical limit when u is very small in comparison to c this expression does resemble the standard classical expression for kinetic energy. So in fact it is this expression of k which matches the classical expression of energy and not the expression for e e is very very different concept which does not exist in classical physics. So these are our new definitions p is equal to mu where I have defined m is equal to this particular quantity e is equal to mc square where again I have defined m is this thing and k is equal to mc square minus m0c square. Now before I try attempt working out certain problems let us use one of the very very important thing the energy and momentum relationship which is very very important because remember whenever we try to work out the collision problem we write momentum conservation we write energy conservation and then we relate energy to the momentum. In a classical standard collision experiment the relationship is very simple e is just turns out to be equal to p square by 2 m. So these two equations one on momentum conservation another on energy conservation can be linked through this particular equation. But now what we will be seeing we have to obtain a similar expression which is valid in relativity and that is what I am going to do just now. This is what I am calling as energy momentum relationship. So I have tried to evaluate what is p square minus e square by c square p square I have just now defined as this particular quantity e square I have defined as m0 gamma u square c actually e square should be c to power 4 but there is a divided by c square. So this remains c square all right the dimension of e by c is the same as the dimension of momentum okay you just take here m0 gamma u square common this becomes minus u square minus c square okay now you write this gamma square as 1 upon under root 1 minus u square by c square do a little bit of manipulation mathematical manipulation you will get that this expression will turn out to be minus m0 square c square. So this is a very very standard relationship which you always remember and use in relativity e square is equal to p square c square plus m0 square c to the power 4 remember this relationship exist is between momentum of a particle and the total relativistic energy many times to differentiate what type of energy we mean this e often we call as the total relativistic energy okay just to differentiate it from total classical energy which is generally written as potential energy plus kinetic energy all right. So this is the relationship which we will be using in some of the problems later sometimes we are also interested in finding out a relationship between kinetic energy and momentum all right so that expression also I am giving but most of the time we tend to use this particular expression but you know in case it is needed we can also use this particular expression the relationship between kinetic energy and momentum okay we have used we have just now derived this equation so I am just taking this p square c square as e square minus m0 square c to the power 4 so this is what I have written here e square e square c square is equal to e square minus m0 square c to the power 4 now e I write as k plus m0 c square for this is the way I had defined the kinetic energy earlier k is equal to this thing so your e will turn out to be k plus m0 c square so k plus m0 c square to the power 2 square so take expand this square so you will get k square plus 2 m0 c square k and then you will get m0 square c to the power 4 which will cancel with this so you will get p square c square is equal to k square plus 2 m0 c square k this is the relationship between the kinetic energy and momentum and this in the classical limit meaning you becoming very small in comparison to c should give you standard definition of kinetic energy is equal to p square by 2 m that is what I am trying to show this was this expression I take k out this is k plus 2 m0 c square now in the limit that kinetic energy is very very small in comparison to 2 m0 c square now you can neglect this particular k and then you will get c square will cancel you will get k is equal to p square by 2 m0 which is the classical expression classical relationship between the momentum and energy so in fact you know whenever we are trying to solve energy momentum problem okay students often ask the question when I am going to use relativistic expression when I am when I am bound to use relativistic expression when I can use classical expressions of course classical expressions are simple enough so if at all I can use classical expression nothing like it but if I cannot use or if I am going to get wrong result by using classical expression I should use relativistic expression now in the simple kinematics it is rather easy to say that whenever the velocities are close to velocity of light in fact I have told you that you know if v is of the order of 0.1c you make only 1% correction okay so you know what are their limits when your v is let us say lying between or the speeds that we are interested are lying between 0.1c and let us say c that time you are probably have to use the classical relativistic expression in terms of energy the limit that we always describe that in order that this particular thing has to maintain k should be much smaller than the rest mass energy so if I am talking of the energy in terms of energy okay I can use classical expression so long my kinetic energy of the particle is much smaller in comparison to it is a rest mass energy just now I have told you that an electron has approximately a rest mass energy m0c square is of the order of half m e b 0.5 m e b okay so if there is a problem which says that one particular electron travels with an energy with a kinetic energy of 1 k e v this 1 k e v is very small in comparison to half m e b okay so I can use the classical expression but if I say that there is a one particular electron which is moving with the speed which has a total kinetic energy of let us say 1 m e b then you are bound to use relativistic expression because this k cannot be neglected in terms of comparison to m0c square so basically the thumb rule is that if your kinetic energies are much smaller than rest mass energy you can go to the classical limit otherwise you must use relativistic expressions now let me repeat in every process both energy and momentum have to be conserved in the completely inelastic collision discussed earlier the rest mass of the final particle will be larger than the sum of the rest masses of the initial particles now I will come back to this particular problem and try to show this particular aspect I use the same example now I work out with the correct expression now I have got my correct expression of momentum I have got now my correct expression of energy and I know that this particular problem cannot just be solved by applying only conservation of momentum so now let us try to solve this problem correctly using the new expressions and then show that both energy and momentum will be conserved in s and s frame of reference so the problem is exactly identical as before only thing I am using now correct equations okay to signify that momentum will be conserved in both s and s frame of reference so again you had similar type of situation before collision a mass going like this mass going like this each speed is 0.6 c and after collision it turns out to be this much okay normally you sign in my finding many of relativity issues we always prefer to use speeds 0.6 c or 0.8 c because the gamma value turned out to be somewhat cleaner so let us just let me just write this particular thing gamma u which is 1 upon under root 1 minus u square by c square if u is equal to 0.6 c okay u square will become 0.36 c square if I substitute it here I get 1 minus this is 0.36 c square will cancel out so you will get 1 minus 0.36 remember gamma u is dimensionless quantity so this becomes 1 divided by under root 0.64 fortunately the under root of this particular thing is fairly clean so this required 1.0.8 which is 1.25 similarly you can show that for u is equal to 0.8 c also the gamma value turns out to be fairly clean so whenever we are trying to emphasize only the concepts okay often we have we use the speeds which are sort of gives you a clean number for gammas okay so that is why 0.6 c is a very standard thing so now I find the momentum of the first particle in S frame itself okay so I know that initial momentum will turn out to be 0 but nevertheless let me write the correct expression for momentum so this is the momentum of the first particle before collision the momentum of this let us suppose this is the rest mass of the particle I am not multiplied by gamma u because this speed is 0.6 c okay so just now I have said that gamma will turn out to be 1.25 for this so this is 1.25 and u is supposed to be the speed of this particular particle which is 0.6 c so this momentum will now be given by 1.25 multiplied by 0.6 m0 c so this gamma introducing is necessary within because of the new definition of momentum remember when I worked out this problem earlier I had not used this term which was not correct because at that time I did not have a new definition of momentum which will turn out to be 0.75 m0 c similarly for the second particle it is the same thing except there is a negative sign here I use the exactly the same expression I will get the momentum to be minus 0.75 m0 c and as you can see that sum of the momentum will be 0 plus 0.75 minus 0.75 will give you 0 so this is expected this is nothing surprising okay but now I have used the correct expression for momentum clearly the final momentum is also 0 because you know of course gamma u will be 1 and u anyway is 0 so the momentum is going to be 0 so therefore you will find out that the momentum of the particle after the of the resultant particle will also be 0 but now energy must also be conserved it is necessary that energy is conserved otherwise I will not be able to maintain the universality of conservation of energy and conservation of momentum which I am illustrating by using this particular problem so now I have to write what is the initial energy of the particle remember in s frame in s I have not yet gone to s frame so the energy will be given by mc square so this is m0 gamma u which I have just now calculated as 1.25 okay so the energy of this particular first particular particle is 1.25 m0 c square similarly the energy of the second particle is also 1.25 m0 c square and you know energy is not a vector quantity so when I take the total sum of the energy both have to be added like classical thing the kinetic energy have to be added there okay kinetic energy cannot be subtracted from each other because they are not vector quantity on the other momentum have to be subtracted because they are in different direction and momentum is a vector quantity all right so according to me now the total energy will turn out to be 2.5 m0 c square so in s frame the particle the observer will now find that the rest mass of the particle is 2.5 m0 see first particle rest mass was m0 second was m0 okay but now the new particle that has come does not have a rest mass of 2 m0 but has 2.5 m0 why it has increased because there is certain amount of energy which has been supplied see remember in classical mechanics the energy was not considered but that was not lost okay but that energy now has resulted in an increase of its rest mass as far as the relativity is concerned because this energy is supposed to take care of all forms of energies so this is 2.5 m0 c square so rest mass of this particle would have gone up okay now hence the final energy of the particle should be same even though the speed of the combined particle is 0 so it means that particular particle has only rest mass energy the resultant particle which has been formed as a result of the collision of these two particles because it is at rest in s frame so whatever this particular particle has the energy must be only its rest mass energy so this m0 is equal to c square hence the rest mass of the combined particle is 2.5 m0 which is what I just now explain where the total energy will turn out to be 2.5 times m0 so mass being instead of 2 m0 has become 2.5 m0 because overall there has been some energy which has been taken by this particular combined mass in addition to the normal standard okay now let us try to go back to s frame of reference and try to see that now with these new ideas the momentum and energy are both conserved okay in fact you will see that if this I have not used this particular term I will not have got momentum conservation again it is necessary to conserve this energy also in order to get momentum conserved in s frame of reference let us see this thing let us go to s frame velocity I have already calculated from the first particle which turns out to be 0 okay and again momentum of this particle will turn out to be 0 because u itself is 0 okay so even if you multiply by gamma u does not make a difference because you multiply it by 0 so you will get it 0 so as far as the first particle is concerned okay in s frame frame like before momentum will turn out to be 0 but now let us look at the second particle for which we had earlier calculated the velocity okay and now I have to calculate the momentum using new definition of momentum this was the velocity that I have obtained now I have to calculate the momentum using this particular definition so what I will have to do I have to not only multiply by m naught but I have to multiply by gamma u and when I calculated gamma u the value of u that I have to take will be this all right so this is what I am doing here if I calculate the momentum I am multiplied by m naught c this was my original velocity okay but I have to multiply it by 1 upon under root 1 minus u square by c square and u is 1.2 upon 1.36 c c square will cancel it out so the second momentum of the second particle in this particular frame of reference will turn out to be this earlier we had used only this particular term so we did not get the momentum conservation now I have to use this particular term all right and there is a negative sign because this particle in s frame of reference of course moves in minus x direction so little bit of mathematics we will find out that this momentum of the second particle will turn out to be minus 1.875 m naught c m naught c has the dimension of momentum okay so this is the momentum of the second particle if I take total initial momentum total initial momentum will be first particle is 0 second particle is minus 1.875 so the total momentum of the system was minus 1.875 m naught c I hope it is clear now let us look whether the momentum is conserved so I consider now the second particle okay where I consider the second particle okay not only I have to you I have already calculated the velocity okay using that I have to calculate gamma u calculate new momentum but the resmas that I have to take now because of energy conservation will not be m naught but will be 2.5 m naught and remember that is necessary in order that I get momentum conservation if I still would have used to m naught I would not have got momentum conservation how did I get 2.5 m naught because I also conserved energy in s frame this I have shown it earlier the velocity of the combined particle after the collision will be minus 0.6 c momentum of the combined particle with 0.6 I will get gamma u which will turn out to be equal to 1.25 this is 1.25 0.6 but now I have to use m naught c okay this is the resmas energy the combined particle which I had obtained from conservation of energy so this I have used 2.5 times m naught I will get minus 1.875 m naught c now momentum will turn out to be conserved but remember this turned out to be conserved even after the new definition of momentum only after I conserved energy and obtained that the new resmas is 2.5 m naught but this is what I am trying to emphasize okay even in a process like classical process in which it was a completely inelastic process both energy and momentum have not to be conserved okay this may result to a different resmas that is a different question now let us satisfy ourselves that energy is also conserved okay not only energy but the momentum also should be conserved initially in the s frame of reference what is the energy first particle was at rest in s frame of reference it has only the resmas energy which is m naught c square second particle was having a speed of 1.2 c 1.2 divided by 1.36 c so using that particular thing I will calculate gamma u so it will be gamma u times m naught c square all right now a total initial energy in s prime not kinetic in a total initial energy of the system in s frame of reference final energy will be now gamma u m naught c square the particle is only it is moving with a speed of minus 0.6 c so I calculate this I have taken as 1.25 this indeed turns out to be 3.125 m naught c square so you can see that the energy is also conserved now let us come to the concept of what we call as a zero resmas particle which I had used when I was talking about the photon in fact I am now completing the circle and coming back to the concept of photon in my last lecture these are the expression of energy and momentum that I have written let us suppose m naught is equal to 0 if m naught is equal to 0 normally energy will be 0 and momentum will be 0 because numerator has 0 okay so if resmas of the particle is 0 then it may not have energy it may not have momentum but looking at these equations you also get into one very very interesting possibility remember I can write e is equal to 0 only of course numerator is 0 if resmas is 0 but denominator should not be 0 then only 0 divided by something which is nonzero is 0 if I am dividing two zeros 0 by 0 it may lead you to a constant value okay because on the other hand I can write this as m naught c square is equal to e multiplied by 1 minus u square by c square if this is 0 then even with a nonzero value of e this equation gets satisfied similarly even with a nonzero value of momentum if denominator also happens to be equal to c then this equation will be satisfied so you want this to be 0 this m naught u you can can be written as p into under root 1 minus u square by c square if denominator turns out to be 0 this equation will be satisfied so if it so happens that this denominator is also 0 then in that case it is possible that the particle could have e and particle could have p it introduces a very very interesting way of looking into the issue that you may have a there may be possibility if resmas is 0 the particle may still possess energy and may still possess momentum but in that case denominator must be 0 which means u must be equal to c it means that particular particle must travel with speed of light okay this is what I have been telling earlier and this is what as you know photon is an example of a particular particle which comes from relativity of course in this particular case because kinetic energy has written as e has written as k plus m naught c square because m naught c square turns out to be 0 you will always get e is equal to k and there is another expression which I have used e square is equal to p square c square plus m naught square c power 4 okay if m naught is 0 in that case relationship between energy and momentum will be just e is equal to p c so momentum will just turn out to be equal to e by c this is what I have told earlier that the momentum of a photon can be written as its energy divided by c I give you an example a particle with 0 resmas with energy h nu and momentum h nu by c okay and light consist of photons. Now let I will work out two problems basically one problem is a typical problem and there is another problem which I am talking which will I talk about Compton effect which I had promised you that when I am doing relativity I will be talking about the Compton effect. So this is one example okay let me before I go to this particular thing let me introduce one thing about which you know a lot of students ask question and there is always a confusion see thing is that when I am taking a massive body and when I am talking trying to say that the two bodies two particles collide the mass of that particular particle will go up because it is completely in elastic collision. Now lot of people ask this particular question what will happen for example if two electrons collide in such a fashion okay will the mass of the electron will go up okay of course when I am talking the resmas of the electron my resmas of the electron will always have a unique value okay you cannot emissage a situation in which two particles two electrons for example if at all could combine together okay forget the repulsive force but forget what repulsive force could combine together or for that matter whatever it is okay they combine together and its resmas becomes different okay in that case if you find there is if there is a conservation of energy and conservation of momentum okay and both cannot be obeyed this particular process will not be possible at all in fact this is one of the example which we always give to the students just to show that if there is a electron which is at rest it cannot absorb a photon because it absorbs a photon both energy and momentum cannot be conserved okay because resmas of the electron will always remain resmas of the electron because electron is not expected to have an internal structure but if I am talking with big particle okay and the two big particles collide then in principle it is possible that resmas was up because these have internal structure it has resmas have gone up because there is some sort of internal energy in the system which is existing for example this could be hot this could have become hot okay when this has become hot actually what particle being hot means okay that no there is more lattice vibrations which are happening in this particular thing so I can emissage a situation when this particular particle can be heated and therefore its resmas can go up but on this other electron I do not expect this to be so because it is not supposed to have an internal structure okay now if you put the electron give them energy only thing that it can do is to translate so only energy that it will be able to have is vanishing of kinetic energy so resmas could not change could not change this is a question which lot of students asked so I thought you know I will sort of emphasize on this particular thing that whenever I am saying that after result of collision the mass has gone up okay these are assuming that these particular particles are massive which have some sort of internal structures so now let us take one particular example which is a particle decay okay now as I say particle mass and massive particle and nonmassive particle you know all are replaced in relativity by just in the form of energies so let us suppose that there is a speed there is a particular particle which has a resmas of 1.35 MeV upon c square see in fact as I said then you know people necessarily express many times express the resmas in the unit of energy that is what I have been done to make it more scientifically correct or physically correct I have said divided by c square because it is mc square which has been given as 135 MeV loosely speaking again we are generally used to lose loose talking in physics also so long we understand it if I could have said that what in the resmas is 135 MeV okay but it is sort of understood that what I meant was mc square is equal to 135 MeV so let me be very clear here that resmas of the particle is 135 MeV by c square it is measured to be 0.8c in a frame s okay the particle is found to decay into two photons so this particle disappears and decays into two photons each of these photon makes an angle theta with the initial direction of the particle find the angle and the frequency of the photon standard applying conservation of energy conservation of momentum so let us suppose we have a frame s and let us suppose we have a particle here which is also traveling with a speed of 0.8c this particular particle decays so the particle goes away and now we have two particles which are coming up one with energy let us say EP1 let us say another EP2 you have to find out what is the angle of this particular thing and you have to find out what will be the frequency of these photons okay simple application of conservation of energy and momentum so what you have to do you have to find out what is the initial energy of this particle you have to find out what is the energy of the initial momentum of the particle then using these theta values and EP1 and EP2 you have to find out what will be the final energies and final momentum of course when you are talking of momentum conservation momentum being vector so x direction and y direction momentum or momentum will be conserved separately so you have to conserve x component of momentum you have to conserve y component of momentum and then solve the equations to find out what will be the value of theta and what will be the value of energies of the photons once we know the energy of the photon I can always find out what will be the frequency of the photon so let us work it out so let us talk about about the initial particle we talk about gamma u again I said 0.8 gives you a clean number there is no under root so this turns out to be 5 by 3 so let us write the energy of the incident particle which will be written by gamma u m0 c square m0 c square I have given you is 135 Mev gamma u turns out to be 5 by 3 it means the total initial energy of the particle is 5 by 3 multiplied by 135 is 225 Mev so the energy of the particle which was just before decay was 225 Mev now what would have been its momentum its momentum would have been again gamma u m0 u gamma u is 5 by 3 m0 is 535 135 by c square multiplied by u which is 0.8 c will you express this particular thing the momentum will be 180 Mev by c often we represent in these particular problems the momentum in the unit of Mev by c so this particular p will be equal to c but the c will cancel out one of the squares so you will have c so this I have written c here if you take all these numbers this number will give you 180 this will be 180 Mev by c this will be the momentum of this particular particle now it decays into photons so I have to conserve energy and momentum I have to look at the energy and momentum of the final particles and then conserve them if e p1 and e p2 are the energies of the two photons then of course I must get because they are not vectors so some of their energy has to be 225 Mev this is my energy conservation because initially the energy of the particle was 225 Mev so as a result of this decay you will find that the final energy has to be 225 Mev now I conserve the momentum along the x direction okay so you initial momentum was only along the x direction but when you are talking the momentum of these two particles okay this will also have a y component of course y component will cancel it out and the x component you have to find it out here so this will be e p1 cos theta plus e p2 cos theta rather not e p1 but p1 cos theta and p1 cos theta again okay and because these are photons so I know relationship between energy and momentum are fairly simple momentum of this particular photon will be just e by c so I have written e by c cos theta then similarly for the second particle e by c cos theta and we originally the momentum was only along the x direction so some of these two momentum must be the momentum in the x direction of the original particle which was 180 Mev upon c so these two must be 180 Mev upon c now in y direction there was no initial momentum so in final direction also there cannot be any momentum therefore the y component of the momentum must be equal and opposite so I have just cancelled this particular thing e p1 upon c which was the momentum value sin theta which is the y component e p2 by c sin theta they must be equal now because you have been told that these thetas are same so these thetas will cancel out c will cancel out which gives you e p1 is equal to e p2 it means you know the particle share equal energy so energy of each particle will be 225 divided by 2 now you put that particular value here these two terms become equal just find out what is the cos theta value so this is what I have done quickly I will just tell you this gives you e p1 is equal to e p2 is equal to 112.5 which is just 225 divided by 2 and cos theta prime because you know you have two times this particular term so when you are taking this you know you will get two times 112.5 which will give you 225 and this c will eventually cancel it out so you will get cos theta is equal to 180 divided by 225 okay just calculate theta from this transfer to be 37 degree approximate so this particular photon will travel at an angle of 37 degrees because it has to conserve both energy and momentum. Now let us come back to the our B particle duality from which we had started our series of lectures at that time we had mentioned about this thing probably but you know they are not discussed we had said that photoelectric effect actually provides a very standard proof of the presence of photons and that fact that photon possess energy which is equal to h nu which combines the both wave and particle aspect okay I do not remember probably had mentioned that particular time that photoelectric effect does not give you any idea about the moment about the photon it only tells about the energy of the photon because as I have told you that if you have really a particular particle let me guess let us assume that there is a particle at rest it does not matter whether if it is moving with the constant velocity also because you can always go to a frame of reference in which this particular particle is at rest okay it can be shown that this particular particle cannot absorb a photon a photon which comes here it is not possible that this particular particle will absorb photon because in that particular case both energy and momentum cannot be conserved okay so this particular process will not take place because I expect conservation and momentum and energy to be always valid if it is not possible to satisfy that particular process is bad it is not possible to have that particular process so if free particle cannot absorb a photon alright so either you have to make photon that this particular particle bound then it is possible to absorb the photon or only part of the energy of this particular photon should be given to this particular particle and the particle should require so first experiment is what we call as a photoelectric effect experiment where it is always performed when this particular particle this particular electron is sort of bound inside the system when it is bound in this particular system it requires certain amount of energy to release the electron then it is possible to conserve energy and momentum but because this particular photon gets now bound to a much bigger atom or a solid okay essentially the momentum conservation is absorbed momentum is absorbed by that particular particle and you are it is very very difficult to test the conservation of momentum it is almost like that that if I throw a ball on a wall in your room okay and I want to test conservation of momentum I will not be successful because to meet up here as a momentum has not conserved because the ball has gone and get reflected back so what was the initial momentum the wall okay has just reverse backwards direction the fact is that momentum is still conserved in that particular process whatever was the difference of the momentum was taken by this wall which has been rigidly connected to earth so strictly speaking the earth velocity would have slightly changed because of this collision all right but on the other hand earth is so big that by making such small change its velocity you will never be able to measure it okay so you will by performing this experiment you will never be able to check whether momentum conservation is valid or not so that is what is happening in the case of photoelectricity experiment that it is not possible to check whether of course you can check that energy is equal to h nu but you will not be able to check that momentum is h nu by c for that I require the second process in which only part of the energy of the photon is taken by the electron and that is what is called Compton effect this is what is Compton effect as I have mentioned now I take electron to be free electron and let us assume that this photon is incident on it but unlike photoelectric effect where I assume that this photon is absorbed here you still have a photon but now with a reduced energy it means with a larger wavelength or a lower frequency so this is a process in which electron interacts with the photon and an outcome there is a photon and of course because part of the energy is given to electron so electron is no longer stationary okay and it also requires it also starts moving so this electron goes this way this photon goes this way and let us suppose these angles are theta and phi I will apply conservation of energy and momentum exactly in the way I applied in a problem which I have done just now so it is a require a photon by a free electron going ahead with the idea that light shows a particle laser course with the photon concept and here I should be able to find out what will be the changes of frequency by application of conservation of energy and momentum and satisfy ourselves and that matches with the experiment so I write conservation equations we can go a little faster now because you know we have solved exactly one problem like this see initial value of the momentum the x direction was h nu by c remember unlike the earlier problem here it was the photon which is coming from the x direction so originally and the electron was at rest so its momentum was 0, photon was h nu upon c finally you have one photon and one electron for photon I can write the momentum as h nu prime by c where nu prime is the new frequency of the photon but about electron I have more complicated relationship because its rest mass is not 0 okay so I have to put p e cos phi then just like I have written earlier in y along the y direction the momentum was 0 initially so finally also it has to be 0 so the y component of the momentum of the photon will be h nu of prime upon c sin theta like the problem I did just now but for the electron now I have to use still p because the relationship between momentum and energy will be given by this expression okay so this is the momentum observation in y direction now conservation of energy even though the electron was at rest it had energy which is the rest mass energy so this was the energy of the electron this was the energy of the photon so overall this was the total energy and that should be conserved now it means that the total energy of the electron okay this includes rest mass energy total relativistic energy of the electron plus h nu prime must match this and of course we have a relationship between e and p which is required you have to solve this equation which is e square is equal to p square c square plus m naught c square c to the power 4 because m naught for electron is not 0 therefore I have to use this particular equation to solve it okay now I will sort of speed through go through faster through the mathematics probably you can work it out yourself I will anyway be uploading these things on model okay so first thing that we do we eliminate phi in fact what we do in actual experiments very very difficult to find out the recoil of the electron and therefore to measure property energy of the electron and moment of the electron or the angle to which electron has required is little tough so I went to want to get rid of this thing so what I will do I will first eliminate phi so for eliminating phi I will use sin phi term and a cosine phi term so I will take sin square phi plus cos square phi which will give you one so I will eliminate first the phi term using these two equations this is what I have done here okay so I have used this particular cosine phi from the first equation sin phi from the second equation okay and this particular cos square phi plus sin square phi is written in this particular fashion so phi gets eliminated in this equation then what we want to do I want to eliminate the momentum of electron also because momentum of electron as I say is difficult to measure that I can do using the fourth equation which is p square c square is equal to e square minus m naught square c to the power 4 I have just changed the terms now as far as this particular energy is concerned this is p square c square I know this energy can be written from this particular expression which is m naught c square plus h nu minus h nu so use that particular expression to write this particular energy okay you work out this particular thing now you can also find out a relationship from this particular thing about momentum so remember this equation I have to write p square c square so p square I will use from this expression so I will multiply by get c square here so this will become h nu minus h nu prime upon cos theta whole square because this will cancel once I multiply by c square similarly this c will cancel because there is their square remember here so for p square c square I can use this expression which I have used here this particular expression then I have used this particular expression for of the square using that particular equation from this equation h nu minus e is equal to m naught c square plus h nu minus h nu prime so once you use this particular expression here okay you will be able to solve these equations and this equation turns out to be somewhat similar simpler if instead of nu I use lambda so I take this c by lambda and eventually I obtain this standard equation of Compton effect that lambda prime minus lambda is equal to h upon m naught c 1 minus cos theta remember theta is the angle to which a photon is scattered okay this is the changed wavelength okay this is what is called a Compton wavelength of electron all right so now let us try to interpret and try to see if we have to perform an experiment how we can perform this particular experiment first let us discuss this particular result we find that the wavelength increases which means energy decreases which is very very standard okay but this increases independent of the wavelength remember increase this depends there is no lambda here appearing here okay so irrespective of what lambda you use you will always get the same thing it depends on h upon m naught c which are all fundamental constants okay and only the angle and depends only on the scattering angle lambda c is equal to h upon m naught c is called Compton wavelength of course in this I have to use mass of electron because I am assuming this particular particle is mass or is electron of course this expression would have been true instead of mass if I would have used some other expression the value of this particular lambda c if you just substitute these things turns out to be 2.43 into 10 power minus 3 nanometers which is exceedingly small now you also notice that delta lambda is equal to 0 for theta is equal to 0 if your theta is equal to 0 then 1 minus 1 will give you 0 so lambda prime minus lambda is 0 and will be equal to 2 lambda c for theta is equal to 180 degrees if I put 180 degrees I get minus 1 so you will get 2 lambda prime 2 times Compton wavelength for theta is equal to 180 degrees therefore because this lambda c being very very small and maximum change that you are going to get is only twice the Compton wavelength okay if you have to take you know difference of like 0 2 or 0 5 angstrom in the wavelength of let us say optical light which has itself a wavelength of 5000 angstrom approximately okay it is very very difficult to measure this thing so these experiments are generally performed using gamma rays okay where wavelength themselves are very very small so the change that you are seeing is a reasonable percentage of the original wavelength and it is easier to determine it okay of course as I said this equation is valid even if you use particle other than electron to scatter the photon of course in that case we have to use m naught as a mass of that particular particle which is scattering the photon now this is experimental setup theoretical it is very easy to say that I assume that there is an electron at rest but how do I put a electron at rest in there in a laboratory it is not possible okay and electrons as you know that are they react very very strongly with the matter and if at all you have to use I mean I mean even the standard photonetary experiment you have to have a vacuum so that you can see those electrons okay or the experiment is performed in a vacuum tube all right so it is not possible the nearest that we can get this free electron is in a metal the professor Suresh must have talked about the metals okay the in metals they are large number of electrons which can be approximately considered free they are not really free in fact free electron 3 does not work but they are the closest to the reality okay so of course they are moving with large speeds but you know as I say I do not mind if the electron moves with a particular speed because I can always go to a frame of reference in which this particular electron is at rest okay so let us assume that in this particular metal we have some electron which are having no force on them so they move with a constant velocity okay though we know that this particular thing is not really correct so we always use a metallic target and allow an incident gamma ray to be incident here then you put your detector at an angle theta from the original direction of gamma rays now only those photons which are scattered in this direction will be detected then you can actually count these particular number of photos for a given number of time given amount of time and plot this as a function of theta by making these measurements as function of theta so this is precisely what we done do in a experiment involving Compton effect when you perform this experiment you get something like this it is a very very interesting you get we see I mean this is not really experimental I have tried to roughly draw it if you want to see some standard curves you know you should probably look into some textbooks okay so the total number of photons which are being detected by a detector in a given time okay as a function of lambda prime okay turned out to be this particular thing or as this is the value of lambda prime which will which are being measured as function you are finding out n as which of this now there are two three important results that you will try to get it here that you find out okay let me also just tell you that what you do as a function of okay let me first explain this thing now you have first of all you find a width which is larger here and a bit which is smaller here okay now we have already done the uncertainty principle so I do not have to explain that there is always going to be a natural line width you know photons will never have exactly the same wavelengths okay there is also always going to be a finite line width but here you find that there is additional broadening of the line width and this particular additional broadening occurs because of the fact that you are actually able to get there are a lot of electrons which are also traveling with different speeds the speed of all the electrons in a metal is not constant because they are traveling with different speeds okay you can show that because of the distribution of the speeds of the electrons you will find that these weights get brought in but what is more important that if you perform okay let me just mention see when you are performed the experiment as a particular theta you also keep on measuring their energy distribution when you are doing theta you are not only counting the total number photons which are coming you are also counting their energies so you can really check that whether their energies or their wavelengths correspond to the expression of the Compton wavelength that I have used now what you do you essentially find at that particular value of theta a large number of photons which are coming with a distribution of energies which are something like this so that is what is the lambda prime for a fixed value of theta now if you perform this experiment at different values of theta what you will find that one of the values which is corresponding to let us say I mean let me put the separation of these particular lines will keep on going up and this particular peak will not change its position only the separation will keep on going up now the second question is that why you are getting two different values of lambda if you take the larger value of lambda prime and calculate it from the original value of lambda you do find that lambda prime and as lambda satisfy that condition of Compton expression that I have obtained but what is interesting that why you are getting also the original wavelength this peak corresponds to approximately the original wavelength and this is present because the fact that inside the metal you do not have only electrons you also have nuclei which are heavy nuclei or heavy basically nuclei or protons or whatever you want to call heavier particles so once the photon comes it need not get scattered only from electrons it can also get scattered from heavier particles like nuclei and they will also get Compton scattered but in that case the shift will be given by h upon m0c where m0 is the mass of that particular heavy particle and m0 being very large the shift Compton shift is going to be exceedingly small therefore you see a peak which corresponds to more or less the original wavelength and there is another peak which results because of actually the Compton scattering of photons from the electrons so you do find that this particular relationship is valid and you prove using this Compton effect experiment that this particular relationship about the momentum of the photon which we have obtained from energy conservation and momentum conservation is experimentally verified. So now I will stop and as I said this will be my last lecture so all the best to you if there is few questions we can probably take. Yes, Namakal, Danmani College of Engineering. Whether the velocity change of two respective objects are involved in a collision will equal or not? See it depends on again the type of initial conditions that you are putting and what are the masses of the particle. It is like a standard collision problem. See like if I am talking of a classical particle, if I am talking of two particular particles which are sort of colliding it depends on for example what is their scattering angle and eventually conservation of energy and momentum. So velocity changes etc would depend on what type of particles are there and how they are interacting. So one can answer this question just in a simple blanket way. It depends on what are the conditions how these particular particles have interacted whether they this is a purely one-dimensional collision or there is a two-dimensional collision and after the collision one particle goes this way another particle goes this way. It depends on so many other factors but only thing that I am emphasizing that irrespective of the type of the collision energy and momentum both have to be conserved. Ranga Swami College. Good morning sir. Good morning. Sir the term mass in special theory of relativity is referred as rest mass. Then how it is possible the massless particles exist here? How is it possible what? Here sir. I did not hear the last part of your question. Massless particles. Okay see I mean as I say I given you my logic about that particular thing that this I mean see only thing when you are saying that there is a massless particle which has a zero rest mass okay this particular particle cannot be put to rest okay because this always move with a speed of light okay and you know in any frame of reference that you go speed of light will always remain speed of light okay so that particular particle you will never be able to put at rest so I mean it is rest mass is zero and therefore you cannot put it at rest. I think that is the best answer I can give. Then the theory of relativity is possible to apply for fundamental particles if it if the mass mass are not considered here most of the particles are not concerned. I am not sure what I am not able to understand your fully question but thing is that of course special theory of relativity is always applied to any physics provided you are talking of that particular energy range and especially in fundamental particles that becomes little more important because when you are talking of the particle interaction you are talking of decays etcetera and all those things okay this particular it is important to talk about the special theory of relativity that is the only we ever can understand them. One more question. Yeah please. Basic difference between special theory of relativity and general theory of relativity. That is what I told last time also in special theory of relativity we are only dealing with the frames of references in which which are inertial. General theory of relativity is more complex in fact I tried to answer this question last time also okay which includes non-inertial frame of reference and try to compare with the gravity take that particular thing especially that is based on the fact that your inertial mass mass and the gravitational mass value turned out to be same okay. So it is little more general which also includes gravity okay while in this particular case of special theory of relativity we are only dealing with inertial frames okay. Yes RBS engineering college. Yeah please. Sir can we see ourselves slower due to real activity, relativity, time relativity. Can we see ourselves slower due to time relativity. See you see the thing relativity is required only when you are changing from one frame to another frame of reference of course let me let let us be very clear. See if I am only my own frame of reference then at that particular time in my frame of reference of course I have to apply new definition of momentum new definition of energy okay but okay if you I am trying to say that my velocity changes you know only in somebody else has to observe. So when I am talking of transformation it is transformation between two different frames okay. So if a particular particle is moving with this particular velocity and is being observed by one observer another observer would always observe this particular particle and may find its velocity to be different. Now in my own frame I am always at rest so there is no way that I can talk of my own velocity I am always at rest in my own frame of reference you have to talk only of different frames and when you are talking of relativity especially these transformation equations these transformation equations talk of what happens in one frame of reference and how do you connect this to what happens in another frame of reference or rather what is observed in another frame of reference okay. Thank you sir welcome sir when light travels in a medium yeah yes sir is there any any equation to show the conservation of energy and momentum. See the thing is that you know I mean see as I say energy and momentum are supposed to be general relationship which is always supposed to be valid okay. Now as far as a light speed of light is travelling in a medium in fact I had told about this particular thing last time also that when I am talking of speed of light being constant in different frame I mean different frames I always mean speed of light in vacuum so I have not been categorically mentioning every time but it is always the speed of light in vacuum when you go into a medium the speed of light will become different but that is what relativity does not talk about it is always about the medium because only in the vacuum that this particular expression c is equal to 1 upon under root epsilon or not is correct. Sir one more question please. Yeah please go ahead. Sir can I have your valuable commands on two paradox is it is it whether the reality in terms of biological age and all that. See the thing is that I have not talked about twin paradox I have not talked about length contraction I would not like to talk about it because that takes enormous amount of time what I will say because they are paradox they are methods of solving those paradoxes you can look at any of the standard books you know in which about they are talking in details about the paradoxes normally at the first year level I do not like to talk about paradoxes they are very very interesting concepts but the thing is that you know that it takes you away from the main physics aspect of relativity and in a limited time we do not want to talk about those things okay but I would suggest that you can look at some of the books you know in which these paradoxes I mean they are three or four paradoxes which we talk in relativity and all of them have been discussed in great detail how to solve these particular things I mean how these paradoxes can be removed I mean can be sort of dissipated with by using proper logics thank you bye bye and have a good lunch