 Hello everybody. So, today we are going to talk of the some other new consequence as well in the sense that new to our course of course of special relativity. We are going to talk of mass energy and momentum in special relativity. I suppose remember how mass was varying with velocity right. So, unlike in classical mechanics where you know if you whatever the mass of a body is if you move if the body moves its mass does not change. But here of course we have this concept of what a rest mass is and the fact that you are in a frame which is moving with some you know some uniform velocity with something else and in a sense the mass itself is moving with certain velocity then you do find the mass of the body has changed. Well we did see some consequences of that today what we are going to do is relate this the relate the mass that is to the total energy and also with the momentum. Now, having said talked of mass and energy I mean it is very certain that we are going to talk of E is equal to m c squared. We are going to spend some time to see how this comes about it is I suppose one of the most well known equations in physics anybody who has even opened a textbook in physics or about read about scientists and what physics is has heard about this equation. So, today what we are going to do is to see a simple way how to derive this from principles which are known to us. So, first you know let us consider a force which is being applied to body and when this force acts what happens is that it displaces the body by an amount by a small amount d x let us say in time d t. So, the small displacement d x and that too in the direction of the force y. So, if you take the dot product you know if you just take you do not need to take the vector part in such f d x and then it also happens in some small time d t. So, what happens I mean what happens when one applies this force well body starts moving I mean if the body was initially at rest and then after the application of the force it retains velocity v let us say small v then what would be the total gain in the kinetic energy of the body well by definition it is the work done is not it I mean. So, that is the energy that the kinetic energy that the body has gained once it reaches a final velocity v from its initial rest position what is that. So, that is integration of f d x of course, well you know in a proper sense one should have taken the vector f dot t x but as we said the displacement is in a direction of the force. So, it hardly matters here. Now, interesting point here is that we write this force now in terms of the rate of change of momentum just see the region next to f d x and you see that it is under the integral sign you have d p d t. So, that p being the momentum gained and d x is the displacement anyway. So, how do you write this momentum you know that momentum is the mass times the velocity. So, we write d of d t of m v and then for d x we just write v d t now why is that. So, you know v being the velocity of the body that is equal to d x by d t and then we have just replaced d x as v d t right. So, how does I mean what does the equation for the expression for the kinetic energy boils down to well it is as we said I mean we are finding the gain in kinetic energy when the particle is you know from has moved from its rest position to its and has attained the final velocity v it is the integral within the limits 0 to v v being the final velocity gained and in the integrand is v of d of m v. Now, check that here we have not talked of whether m is constant or not. So, this form of the kinetic energy is very standard. We can use it for example, let us why do not we use it to find the non relativistic non relativistic expression for the kinetic energy that will clarify the situation a bit. So, for the non relativistic case what do we do here we know that you know mass of a body is constant I mean it is a fixed thing. So, we can take it out of the differential d of m v take it out of the integral sign. So, that the integration of for the kinetic energy is simply v d v and then from 0 to the final velocity detain. So, very simple integration to do the integration the integral rather becomes v square by 2 you know v d v 0 to v that becomes v square by 2. So, the total kinetic energy is nothing but half m v square now this is a known expression to us we know this expression half m v square it is it is used almost everywhere in classical physics classical mechanics by the way. So, of course, what have we done here we have taken mass to be a constant thing. So, as expected now what happens to the relativistic case in the relativistic case remember that the mass is indeed a function of velocity. So, if the mass of course, moves if the body moves with a certain velocity with a large velocity we do see actually you know large is compared to what I mean if it is of course, all velocities we are we will be considering is less than the speed of light. But then if it is approaching the speed of light we do see that the mass has increased by a lot. Of course, I mean when very low speeds it does not matter we are still in the non relativistic limit. But then we have appreciable speeds then we do start feeling that the mass of the body has increased. So, what happens if we consider that. So, the mass of a body being a function of a velocity now then we can no longer take the mass of the m out of the differential and then we have to use the entire thing. Now what would be the expression you are going to put in the exact value of the momentum or the exact value of the mass here that is m 0 divided by root over of 1 minus v square by c square of course, the v is there. So, the kinetic and what is m 0 here m 0 is the rest mass that is the mass of the body that is measured when the body is at rest. So, what is the expression for the kinetic energy then well it is v times d of m 0 v divided by root over of 1 minus v square by c square. Now this might look like a bit complicated, but we do have a thing called integration by parts in mathematics and so I think it is a proper case to utilize such a trick. So, what is this integration by parts. So, it is that if you have an integration of x dy and then you can write this as you know and then y is some complicated function that you have. So, it is nothing but x y minus of integral of y dx. Now what we do is that we identify x as the velocity here and then for y we take you know it is m times that is the momentum. So, m times v that is nothing but m 0 v divided by root over of 1 minus v square by c square. So, let us just do this integration and then get the result. If you are interested in the algebra of it I am going to spend one more slide on it. So, just to show you some of the steps. So, one or two slides that is. So, the integration that we are going to do is one written at the bottom of your screen. So, that is the expression for the relativistic case when mass is a function of velocity. So, the kinetic energy happens to be m 0 v square divided by 1 minus v square by c square root of that minus of m 0 and then we take the integral of v dv divided by root over of 1 minus v square by c square. So, if you do this integral and then put in the proper limits you know just look at the second line. So, when you do the integral and put in the proper limits. So, you get the one that is given in the brackets there that is m 0 c square divided by into root over of 1 minus v square by c square and then the limits from 0 to v. And then the whole expression simplifies. So, whole expression for the kinetic energy simplifies into two parts. See that it simplifies into an expression for the mass of a body which is moving with a certain velocity v times c square. Why the mass of the body being m 0 divided by root over of 1 minus v square by c square that is the total mass of a body when it is moving at certain velocity v that is multiplied by c square minus then m 0 m 0 is what m 0 is just a rest mass of the body times c square c being the velocity of the body. So, that is what we have that is the total kinetic energy of the body. Now, kinetic energy of the body being that we can interpret the total energy of the body as something like m 0 divided by root over 1 minus v square by c square times the c square. So, that will be then equal to m 0 c square. We call it as the rest mass energy if we consider m c square as the total energy plus the kinetic energy. If that sounds a bit confusing it might a little bit. Let us just have a look at the next slide and that will clarify our concept. What is it? So, at rest when the body is at rest we know the kinetic energy is 0, but what happens to the total energy? Remember if you go back to our whole slide, if you go back to our whole slide we see that when the body is at rest the total kinetic energy is 0, but the total energy is still m 0 c square when the body is at rest. So, that justifies the word that justifies the term rest mass energy that is the intrinsic energy associated with the body even when it is at rest. So, when the body starts moving the total energy would be this intrinsic rest mass energy plus whatever kinetic energy the body has gained. So, that will be then the total energy. So, that is exactly what we meant by saying that we interpret the total energy as m c square and m being m 0 divided by 1 minus v square by c square. So, having known this concept of what a rest mass energy is let us just write that as e 0 just to signify that it is you know the rest frame of the body or the energy of the body when the body is at rest. So, as you said what is the total energy of the body when the object is moving it will be e and that will be equal to e 0 times plus the kinetic energy that is and what is that that is nothing but m 0 root over 1 minus v square by c square times c square and what is that that is simply m c square where m is the relativistic mass. So, you see how this e equal to m c square has come about. Now, of course, this is a profound result I mean it has not only altered physics it has altered the world actually. So, it is the first time you are seeing that you know the intrinsic mass is being related with the energy of the body and it has immense consequences in fields of in other branches of physics. For example, if we consider the energy production in the sun is there energies being produced by fusing hydrogen into helium. So, when you so if you add up the masses of hydrogen and you add up the masses of helium you would find that in the final product the total mass is not the same I mean the whatever you started total amount of hydrogen you know the total mass of the hydrogen you started and the total mass of helium that is produced you will see that it is less. So, where has this mass gone into and this mass has been converted into energy. So, hydrogen is being fused into helium and then some amount of energy is released. And that is the and that is the source of that well that is more or less the you know the source of energy production in the sun and to just to give you an example you know just to feel of the numbers or the amount of energy that is being converted from matter per second. About 4.4 into 10 to the power 9 kilograms of matter is being converted to energy per second in the sun. So, how you are going to find the total energy released per second that is the total power actually. So, you are going to multiply this number 4.4 into 10 to the power 9 kg per second into 3 into 10 to the power 8 meter per second square meter per second that is I am sorry meter per second and take the square of that and what do you find you would find that the total energy is nothing but 4 into 10 to the power 26 joule per second. Now, if you just consider the 1000 watt bulb, I remember nobody uses 1000 watt bulbs in the homes anymore you would rather use more efficient sources. But let us you know how how bright a 1000 watt bulb can be ok that is 10 to the power 3 ok that is 10 to the power 3 watts. If you divide 4 into 10 to the power 26 by 20 power 3 it is not it is nothing but 4 into 10 to the power 23. So, per second there the sun is shining like 4 into 10 to the power 23 1000 watt bulbs just have a feel and I just have a feeling of how large this power is. Well actually this is also an important lesson for us to utilize the solar energy in our daily lives. I mean it will you know if you work more on research in solar energy and try to harness this energy which is just coming to us free from the sun lots of this energy problem will be solved. Ok. So, I think you know appreciate how this E equal to MC square comes about and what the consequences of it are ok. Of course, I did not so far talk about nuclear fusion of you know in a very direct way I did not talk of nuclear fusion. But then I give you an example of fusion of hydrogen to helium in stars ok. But some other applications in nuclear physics let us see if you can do such things a little bit later. Ok. So, having talked of energy and mass let us see what the expression for the kinetic energy I mean whatever things that we have discussed so far whether it actually boils down to our known relativistic non relativistic formulae if we take velocities less much much less than the velocity of light. So, let us check if the mass times the energy you know mass times the speed of light squared from where from you know we have derived the kinetic energy let us see whether we can get the proper limit here. Ok. So, we know the expression for the kinetic energy that is the total energy minus the rest mass energy of a body ok. Now since we know that the velocity that we are considering here is much much less than the velocity of light. So, we can we do an approximation here we do a binomial expansion of 1 minus v square by c square to the power minus half here ok that is the denominator of m 0 c square divided by root over 1 minus v square by c square what is that? Well the first term is 1 plus you know the term of importance half times v square by c square. So, you plug this in the expression for the kinetic energy ok it becomes 1 plus half v square by c square times m 0 c square of course and then you subtract m 0 c square from here ok what do you get? It is very simple you get half m 0 v square ok. So, indeed since the mass of the body does not change in non relativistic classical physics it is half m v square and then m being the rest mass of the body ok. So, we do see that it reduces to the classical limit at low speeds. So, having talked of mass energy and then seen and then having seen that it actually does give us the proper limit of the kinetic energy at low speeds. Let us now move a bit and bring in the momentum ok remember we started this module by talking of mass energy and momentum. Let us talk of energy and momentum ok. So, how are they related? Well you know from your previous slides here that the total energy is m 0 divided by 1 minus v square by c square times c square that is m c square. What is the momentum? Well that is m times v m being your familiar m 0 divided by 1 minus v square by c square ok. Well, but looking at these two expressions mathematically you see that you know the mass portion remains the same, but then you know the only thing different is the c square for the energy part and the velocity v for the momentum part. So, what will happen if we you know try to put them together on the same footing? Ok. What happens? For example, if we just do e square minus p square c square. So, what we do is that we square the energy and then from that we subtract p square times c square ok. Now, you will check that of course, that will have the that will have the dimension of the energy square ok. But what is it? Well does it give us something very interesting? Actually it does. I mean if you do the algebra it is rather one or two line algebra. You will see that it is nothing, but when you do e square that is the total energy and from that you subtract p square that is the momentum square times the velocity of light square. You are going to get something very interesting. You are going to get m 0 c square whole of that square of that whole square of that that is. So, that is nothing, but the rest mass squared of the body. So, if the body has a total energy e and it is also having total momentum p I mean it is moving with a certain momentum p then e square minus p square c square is nothing, but the rest mass square of the body. It is an interesting concept actually it is a because what it tells you is that this group of variables e square minus p square c square remains invariant in all reference frames. Of course, these reference frames are moving with constant velocity with respect to each other. So, that is what we have we have e square minus p square c square to remain invariant. Now, well that was that was that was that has the dimension of energy square. Actually it need not be I mean we were also having momentum. So, we could also in a sense have this as the dimension of momentum square. Well, that is very simple all you do is that you divide the entire stuff by c square and then you have e square by c square minus p square. And then the invariant quantity will come out to be m 0 squared c square. Now, remember m 0 is the rest mass of the body and c of course is the velocity of light. And this has the dimension of m 0 c that is has the dimension of momentum and then m 0 square c square has the dimension of momentum square. So, we will spend some time on this relation and then find out certain interesting things. But before that let us continue a bit on these units. Remember in relativity and in like other domains of physics we are quite interested in units in which you measure everything in terms of something in terms of the units of energy. In atomic physics for example, the unit of energy is generally electron gold Euclid physics MeV million electron. So, what is that what is E V? So, E V is nothing but the energy gained by an electron accelerated through a potential difference of 1 volt. So, if you take a million times of this that is 10 to the power 6 E V that gives you MeV and then when you take 10 to the power 9 that is 1 billion E V you get 1 G E V that is a giga electron volt that being the unit of energy. What about the mass? Well normally you would measure everything in grams or kilograms kilograms it is SI unit. So, however if you consider the expression E is equal to m c square you can very well measure mass in terms of energy by c square then you can measure mass everything in terms of in example like MeV by c square or E V by c square G E V by c square. So, that actually helps a lot. So, what about the what about the unit for the momentum mass times the velocity. If you measure velocity in terms of the how much of it is a percentage of velocity of light so if it is MeV by c square then you multiply by c it should be MeV by c or something like energy by c. Well that is very normal because if you just consider E square is equal to p square c square plus m 0 square c to the power 4 then the momentum does turn out to be energy by c. And then you can measure it in terms of MeV by c or let us say the G E V by c if it is an ultra relativistic momentum that you are considering. Now, let us consider something else one other application it is whether one can have massless particles. And I see this as an application of relativity is going to realize that in classical mechanics we kind of never ask such a question. Why? Because well once you have a massless particle or you do not have you have matter which is massless well in classical mechanics such a thing does not exist at all because you see if you have a something massless I mean what is massless in classical mechanics nothing I mean you consider unless you know you have an object which is a material matter then you say it is massless of course then object is not there. So, but in any case if you want to take this argument a little bit further if m is 0 the momentum is 0 the energy is 0. So, such a question does not happen does not arise at all. So, massless particle cannot exist in classical physics or in a non relativistic situation. However, in relativistic situation do we have a massless particle to exist does it you know well it is more like in classical physics it forbids the existence of massless particle, but thus relativity forbids such an existence. Let us let us just check it out well if indeed the rest mass is 0 you would find the total mass to be 0 if velocity is less than c why because m 0 is 0 and the denominator is 1 minus v square by c square is finite and then that is so the total mass m is 0 and then immediately you are going to figure out that the momentum is 0 and the energy of course is 0 fine. So, if the rest mass of a body is indeed 0 and it is moving with a velocity less than the velocity of light then massless particle should not exist. But what happens if we still have the rest mass to be 0, but the particle is moving with the velocity of light itself it is strange is it it. So, then what you see is that the expression for the momentum and the energy becomes 0 by 0 it is an indeterminate form it is an indeterminate you cannot determine it is 0 by 0 form. So, it is basically you cannot say that it is 0 neither can you say it is actually infinite it is actually indeterminate it can have any value. So, in a sense relativity does not forbid the existence of massless particle if and in this case and only if it moves with the velocity of light. So, that is a very important conclusion that you can draw it is that massless particle will be existing only and they will have energy and momentum provided the travel with the speed of light. Actually can you can you can you name such a thing exactly you can it is called the photo. It is the quanta of light then how are the energy and momentum related in this case? Well, since the rest mass is 0 then energy is equal to p times the velocity of light here it is e times p because otherwise e square is equal to p square c square plus m naught square c per 4. So, the m naught square being 0. So, you have e square is equal to p square c square and here. So, e is equal to p into c that is one. So, that is an important conclusion that we have special relativity. So, now let us change tack a little bit and return back to our energy momentum relation and see something quite interesting. So, what do we have here? We have basically e square by c square minus the momentum square. Remember this what we write as the small p that is the 3 momentum that is equal to m 0 square c square and then I write that as capital p square just to give you an idea that the invariant quantity that you have has a dimension of momentum. What kind of momentum is that? It is definitely not the 3 vector momentum that we do in our daily lives. So, that is what I said. So, the capital p here has the dimension of momentum and then p square of course, has the dimension of momentum squared here which is different from small p. Now, what is small p? Now, small p is let us say it is the momentum well it is the 3 momentum of a particle. Now, if you take the 3 momentum squared, so the norm of the momentum I put a vector sign on top of small p. What does it mean? It means that you take a dot product p dot p small p dot small p. Then what does it turn out to be? It is nothing but p 1 square plus p 2 square plus p 3 square. If I write for example, p x square plus p y square plus p z square in Cartesian coordinates and that is invariant, you know that because that is equal to the length of a vector that is invariant. Even if you rotate the vector, the length remains the same. So, the vector, the 3 vector is the one in which we put the vector sign and then we have 3 components. So, we have 3 independent components. Now, check that the momentum or the you know we can define a 4 momentum which capital P whose which has of course, 4 components and to keep in you know in in consonance with the 3 momentum that we had earlier p 1, p 2, p 3 which is denoted by vector p. We have the first part as p 0. We just call this as the 0th component of the 4 momentum and then if we define something like the 4 momentum dot product, I put the dot product in inverted commas and then you put a bigger dot p big dot p and all capital P's. Then we define that as p 0 square minus p dot p the vector p dot vector p, then we can get something quite interesting. Actually, we can relate what we have found here with the energy momentum in special relativity. So, if you identify p 0 as e by c and so the total p becomes the capital P becomes e by c and then the other vector is the 3 vector. Then the you know the 4 vector dot product becomes e square by c square minus small p square and we know that that is invariant. Now, just like the dot product of a vector was indeed invariant here too, the dot product or the big dot product of the 4 vector is also invariant. So, there is something which is happening here. I mean, so I think if you have not got it in these two slides, do not worry. I am going to talk a little bit more about the 4 vectors and space time a little bit later. So, I just want to tell you that there is something interesting that you can relate this energy momentum with things called the 4 vectors. Let us see what that is. Well, to change stack yet again and do a little bit of formal stuff in terms of Minkowski, we are doing relativity, Wanda's relativity in terms of Minkowski space time remember in special relativity. So, it is not only the space part, but we also have the time part which is which is equally important here. So, because here time is no longer an independent quantity, it also depends on the space quantities. That is what we saw in our Lordin's transformations. So, if you define a quantity, an event, let us say or the world point by 4 space time coordinates, since the velocity of light is C which is constant in all frames. So, what we have here is that we take X as CT and then XYZ. So, why have we taken CT? So, that we preserve the same dimension as the space coordinates or you could even have make it look like you just say you have 4 coordinates X0, X1, X2, X3 and then X0 is the 0th in a sense, 0th component of the thing called X capital X here and then small X1, X2, X3 these are the space coordinates. So, that is what we say X0 is equal to CT and then the vector X is X1, X2, X3 or XYZ if you wish in Cartesian coordinates. Now, what is the squared norm of this quantity? As we have defined, it is C squared T squared minus X1 squared minus X2 squared minus X3 squared. So, it is nothing but C squared T squared minus vector small x dot vector small x. So, what we have? So, what we have is that in S frame if we have a point which is denoted by you know one of this world space time events denoted by CT XYZ. So, certain event is occurring at a certain time in a certain position space. Now, the same event will be viewed by someone in the S frame which is moving with velocity V with respect to the S frame as CT prime and at time T prime that is and then at position X prime, Y prime, Z prime. Remember at T equal to and T equal to T prime is equal to 0, this two frames were coinciding and then it started with the move with velocity uniform velocity V that is. So, how are these things related? So, these quantities are actually related by Lorentz transformations. So, what is X prime? Well, X prime is nothing but I mean if you use the notations gamma and beta remember we had used we are using them so that the expressions become less cumbersome. So, beta beings V by C and then gamma becomes 1 minus beta square root over or to the power minus half basically 1 by root over 1 minus V square by C square. Since we are moving uniformly along S frame is moving uniformly along the common X axis and so what is X prime? So, X prime is nothing but in terms of the X quantities, it is gamma times X minus beta CT and Y prime of course is equal to Y and then Z prime is Z and then CT prime is gamma CT minus beta X. So, what happens to the quantity capital X prime? So, if you do C square T square minus X square minus Y square minus Z square you should check that this becomes equal to C square T prime square minus X prime square minus Y prime square minus Z prime square. So, the norm of the quantity X capital X that is of the 4 vector remains invariant whether you are doing it in the S frame or the S prime frame. So, that is what we have and in fact that gives us the idea that if we have, if we can figure out a set of 4 quantities of 4 numbers that transform as the components of the 4 vector X capital X that is they will be 4 vectors. So, and then not only that I mean just about any 4 quantities will not be any 4 numbers you just club them together they will not be 4 vectors. Well their components should transform like the components of the space-time coordinate capital X which is basically they should transform by the same Lorentz transformation. Not only that the norm of the 4 vector that one has constructed should be invariant under any frame. To give an example just think of any other 4 vector let us say a general 4 vector some vector A remember since I am using the word 4 vector not a 3 vector that is what I have not put a vector sign on top of A capital A that is. So, in the S frame that is it has components A0, AX, AY, AZ and in the S frame it has components A0 prime, AX prime, AY prime and AZ prime and that is the vector A prime that is the 4 vector A prime. Now, if A is a 4 vector in S frame and S prime is the 4 vector seen from rather A prime is the same 4 vector seen from the S prime frame then there are certain conditions that should be obeyed. Well first of all the components of A and the components of A prime should be related by the Lorentz transformation. And not only that the squared norm of A and A prime they should be equal and it is actually Lorentz invariant. So, that is exactly what we had for our energy and momentum if we define the 4 momentum as E by C and that is the 0th part and then the other 3 parts are the familiar 3 momentum vectors let us say the 3 linear momentum PXP by PZ. So, if capital P that is the 4 momentum vector in S frame which has components E by C and PXP by PZ and then the same thing seen from the prime frame is capital P prime and that is E prime by C PX prime PY prime and PZ prime small PX prime PY prime PZ prime that is the 3 momentum vector in the prime frame. Then the components of P and P prime are related by none other than the Lorentz transformations. Well this has some immense consequences actually because in a sense this will help us in relating the energy and momentum in one frame with the energy and momentum in another frame. And not only that you can also you can well you can also relate energy momentum conservation directly in the S frame and the S prime frame. So, more of that a little later but then to summarize what we have talked of 3 vector and the 4 vector let us just spend 2 more slides on it. For a 3 vector let us say the velocity vector the radial vector things like that 3 vector of course has 3 components. So, we put a vector sign on top of that and 4 vector of course will have 4 components. Now why is it that we call this 3 component thing as a vector by the way? Well it is because if you are changing your coordinate system and then you are following you know the change of coordinate system is due to is the change of coordinate system is due to this transformation you need this transformation matrix R to go from one coordinate system to another. And similarly in this case of 4 vector it is the Lorentz transformation which is transforming from S to S prime. Now a vector I mean we define a vector to be that to be that physical entity is that look at the 3 vector part. So, if you look at 3 vector part and then you see that x vector x is transformed to vector x prime. Now it is the same transformation matrix which will do the transformation. So, the transformation matrix which is transforming coordinate system O to O prime the same transformation matrix will be responsible it is the same transformation matrix which will transform x to x prime. So, that is a vector and not only that the length of this vector in the O system and the length of the vector in the O prime system will be preserved. Similarly, what you see is that it is a Lorentz transformation which is transforming the quantities in the S frame to the S prime frame. So, it is the same Lorentz transformation which is going to transform the components of the 4 vector in S frame the components of the 4 vector in the S prime frame just like the transformation matrix R transform the components of the vector x in O system to the trans components of the vector x prime in the O prime system. And just like the length of the 3 vector was constant here what you see that the square norm is Lorentz invariant. So, that gives you an idea why we had used this word vector for this 4 component entity in special relativity. Now, of course, there are lots of applications to using 4 vectors. We have just been introduced to two of the 4 vectors one of course is the space time 4 vector and another is the 4 momentum 4 vector. Now, these are going to have applications in relativistic kinematics and things like relativistic collisions. So, let us talk a bit about such a thing about our coordinate systems in a case where such collisions can happen namely the laboratory and the center of mass system. So, what is it? So, basically we are considering the collision of two particles A and B let us say there are certain there are rest masses M A and M B. And let us say that well you can look at this either in the laboratory system and the center of mass system. Well, in case we want to know what these systems are the lab system is one in which one of the particles is at rest and the other comes and hits it with a certain momentum you know certain momentum let us say 3 momentum P or vector P. In this case, let particle B be at rest and then particle A is moving with a certain momentum in the lab frame P AL notice the vector sign and then the subscript AL L stands for the laboratory quantity. So, wherever you see this L it means that it is the lab quantity. So, in a sense what is the momentum for vectors here? Well capital P AL that is the we will read 4 quantities the first quantity relates to the energy and well it is basically the dimension of momentum it is but involves the energy that is E AL by C and then the 3 vector with which or the momentum with which the particle A is moving. What about P BL that is the 4 momentum vector for particle B in the laboratory frame well that is E BL by C and then remember it is at rest. So, it is total moment that is a linear momentum is 0. Now, what is E AL? E AL you know that it is MA into C square and then what about E BL? E BL is nothing but MB into C square. Now remember here E AL now when I write MA here it is MA divided by 1 minus V square by C square actually that is assumed within when I write E AL here. But when I write E BL here remember that it is just the rest mass squared. So, we conclude our discussions here today once we have seen the differences between the 3 vector and the 4 vector now we realize why we call it as a 4 vector itself keeping it you know some sort of consonants with this component of a 3 vector. We are going to see in our next topic the applications of this of the 4 vector in in special relativity especially in relativistic collisions and we will also do some more problems in relativistic kinematics in our next discussion. Thank you very much.