 Hello everybody. So, today we will be talking of another consequence of special relativity. So, we will do some topics in relativistic kinematics. And if one recalls we had to redefine mass, we had this famous mass energy equivalence special relativity gives us that. We also had to redefine mass. So, as to preserve the conservation of momentum, conservation of momentum in relativity. And talking of both energy and momentum how are they related? We know that E is equal to m c square and then this m or the total mass of body that is actually related with speed at which this body is moving ok. So, if m 0 which you see here is the rest mass of a body then if it is moving at a certain velocity v its mass becomes m divided by root over of 1 minus v square by c square ok fine. And consequently we also have the momentum defined as the mass multiplied by the velocity, but then the mass again here is velocity dependent ok. So, how about the relation between them? Why do not we square them up and you know if we square the energy and then multiply a c square to the momentum do a bit of algebra it turns out and then we subtract p square c square from the square of the energy. It turns out that we get something which is which is Lorentz invariant or in other words it does not change when you change frames or it does not change in frames which are moving with a certain constant velocity with respect to each other. So, you actually get the m 0 c square and the m 0 being the rest mass of a body and then c square being the velocity of the c being the velocity of light here ok. That is pretty interesting. So, if we now write it in a certain way actually if we write all these things in the dimension of momentum square we get a nice relation like you get a e square that is a total energy square divided by the velocity of light square that gets the that is get the dimension of that gets the dimension of momentum square that is and that is the linear momentum square that we are talking of here and then you subtract the momentum which I denote by small p you get something which is invariant ok. Now, this I denote by something like the capital p squared and just make a comment that still this is momentum square that is all. Let us see is there something behind all these things ok. Now, to check that what is there behind we if you recall what was a 3 momentum square ok. So, if you know a 3 momentum square think of the thing in the Cartesian system or in any Cartesian ok the Cartesian system. So, what you see the 3 momentum square you get the x component square the y component square and the z component square or component 1, 2, 3 squared each they add them up we are invariant quantity and you all know that this momentum is a vector it has 3 components. Similarly, the thing that we talk of when we talk of a 4 momentum we have we can also talk of something called a 4 momentum square and then what do we get and if you define the dot product of or we define a dot product for the case of 4 momentum as you know the initial the 0th component we call one of the component the first component with 0th component and the other 3 components which are equivalent to the usual the vector in the 3 dimension that we all know of and then we define the 4 momentum the square of the 4 momentum with the help of the special dot product that is why when I say a dot product I do not put a simple dot, but a rather bigger dot ok. So, that is equal to actually p 0 square minus the 3 dot product of the momentum the linear momentum that is here what we have here ok. Now, why do we call it as the 4 momentum more of it later, but here what we see is that if we identify this p 0 of the 0th component of the 4 momentum as e by c and then the other 3 components as the linear momentum itself then what we get is the dot product of this on this 4 momentum dot product shown by p capital p dot capital p is still invariant quantity ok. Now, the notation that I will be using is for a small p will always be the 3 momentum if I put a vector you also know that it is a 3 momentum vector, but for a 4 momentum vector not put any vectors on top of it and not only that I will always use capital letters for it. So, capital p is a 4 momentum and then the small p is the 3 momentum that we talk of here, but it is also interesting that we know that the 3 momentum square that is an invariant quantity ok, but you know the dot product or when you take the square of a vector you get the length of a vector which is which is scalar. So, that is an invariant quantity. Similarly, what we see here is that when we take the dot product of the 4 momentum with itself we also get an invariant quantity ok that something to think about ok. Now, let us recall a few more things about the 4 momentum which we may have covered earlier, but given today's topic of relativistic kinematics I thought it will be useful for us to view a bit of the 4 momentum formalism again a bit ok. So, just to recall we can also think of the relativistic phenomenon in the Minkowski space time ok, where an event is given by an event or a world point is specified by 4 quantities, 3 space quantities, 1 time coordinate ok. Of course, you can multiply the constant velocity of light to time so that you have the dimension of length for all the coordinates ok. And then that is what we do we can simply multiply c to the time coordinate and then we have x, y, z let us say. So, the 3 space coordinates and then we have the same dimension for all and then we can simply write x0 to be the 0th component and the other 3 components for the other 3, the 3 vector here ok. Now, it is also interesting when I take the squared norm or then the 4 dot product of x with itself I will get c square t square minus x dot x like the 4 dot product we had done earlier for the momentum. And how does the components change when we go from one system to another let us say we go from the S frame where you have the coordinate ct at a certain time and then x, y, z and then the same event if it is measured with the help of from S prime frame. And then you think of that frame as and you measure it at a certain time t prime and x prime, y prime, z prime. Of course, you now know that these are related by Lorentz transformations ok. So, the components of this 4 vector is actually related by Lorentz transformations which we have covered earlier. And it is ok just to recall let us read the first line. So, it is the how is the x component in the S prime frame related with the x component with the S frame. You see that let us x prime is actually equal to gamma times x minus beta ct ok. So, where beta is the velocity with which the S prime frame is moving divided by the speed of light ok. And then the gamma what you see here is nothing but the 1 by the square root of 1 minus beta square ok. So, what is also interesting is that if one takes the 4 dot product the 4 dot product in the S frame and the S prime frame can easily check that its norm is equal ok. That is again something we keep in mind it is that the norm of a 4 vector is is becoming in what we check is of for this is becoming invariant when when is actually invariant when you go from one frame to another ok. Now, just as an aside we can also recall that the norm of a 3 vector is also invariant when you change frames ok ok. So, to make things a little bit more general we we call all quantities or rather all quantities which has 4 sets of sets of numbers here. Now, and if they transform and if they transform in the same way as as Lorentz transformations and not only that that the squared norm of of of such a quantity a is become become same whether you are in S frame or the S prime frame then such a thing are actually 4 vectors ok. And to summarize what we have is that we have a 4 if a is a 4 vector in S frame then in the S prime frame the same 4 vector will have of course, the same norm, but its components will change ok, but but the components will change in accordance with what with the accordance with the Lorentz transformations ok. So, that is what we have the components of a 4 vector will change you know the components are related by the Lorentz transformations and then the the norm is invariant fine. So, just to put things in perspective again if we talk of a 4 vector the momentum 4 vector here ok. So, the 0th component is E by C that is the energy by C and then P x, P y, P z are the other 3 components and in the S prime frame which is moving with a certain constant velocity v with respect to the S frame the the components will be different ok of course, it will be different, but the norms of the 4 vectors would be the same ok. So, now you know just as an aside why do we call it as a 4 vector ok. Recall what we have in a 3 dimension space ok when we have a vector. So, what is a vector? So, if we recall what a vector is it is actually that mathematical physical entity which has 3 components here as a 3 vector that is why it is called a 3 vector. Now a vector what is the special property of it? It is that if one changes goes from one coordinate system to another now the transformation matrix which one has to use to go from one frame to another ok that same transformation matrix is going to change the same vector which is in the old system to the new system. For example, here in this example if x is a vector in O system and x prime is a vector in the O prime system. See the transformation matrix which is transforming O to O prime will be the same which transforms x to x prime and not only that the length of this vector is actually the same that is a scalar ok. So, that does not change when you change a coordinate systems. If you recall what we have just talked of this quote unquote 4 vector see some amazing properties which are quite similar with the 3 vector. First of all it is called 4 components and then the components of the you know when you go from one system to another the what is the transformation matrix here it is the Lorentz transformations. And it is the same Lorentz transformations which is transforming the components of this quantity x or this 4 vector from one system from one frame to another and not only that its norm is fixed ok. So, that of course, justifies the word the vector here what you have used and also it has 4 components that is what you call as a 4 vector ok. So, with this let us with this background now I think we can go over to an application of relativistic kinematics ok where let us take a simple example let us study two body collisions ok. Now as in classical mechanics we can study this in the lab system and also in the center of mass system ok. So, what is a lab system when we consider the collision of two bodies ok. So, the lab system is one in which one of the bodies is at rest and the other comes and hits it ok. To keep things simple I mean we just take the direction of motion of one of these bodies of one which is moving to be along the x axis ok and it is on a plane let us say ok. Anyway we also take one of the bodies at rest. So, let us just take the body B at rest and the body A which is coming with a certain linear momentum P A and if you see we have also put a subscript L and also a vector. So, that is an indication P A L. So, that is the indication the P has a vector. So, that is a 3 vector and it is related with the first body A and then it is in the lab system. So, that is the quantity L that is the reason I put this quantity L. Now what would be the 4 vector for such a for these bodies ok if recall that the 0th component of the 4 vector relates with the energy ok. So, we have the energy of particle A in or the body A in lab system divided by C that is the speed of light and then the other 3 components are the 3 vectors or the 3 components of a 3 vector and what is E A L. So, that is nothing, but M A into C square ok that is the mass of A and then what about P B L that is the 4 vector P B L. So, recall that again it is at rest. So, the 3 momentum part of it is 0 ok, but it is interesting that its energy is not 0 why because of course, it has rest mass and then you have rest mass energy. So, you have M B into C square that will be the rest mass energy now if you take M B C square divided by C you get M B C and so, you get the 4 vector P B L that is capital P B L to be M B C and then the since the 3 vector part is 0. So, we have it 0 ok. So, there is of course, another system in which we go over to the center of mass of the system. So, all the measurements are being made from the center of mass of the system here and in this particular case we have the case in which the bodies are moving in a such a way that the total linear momentum turns out to be 0 here ok. So, the same collision process you are looking at a frame you are looking in a way in which the both are moving towards each other in a such a way that the sum of the linear momentum are 0 ok. And it is easy how to write the momentum 4 vectors here for P A notice that I have not used any subscript L here. So, except maybe when I only the subscript for the bodies or the particles have been used here. So, the other subscript the C M is not required when superfluous thing. So, if I do not write it it means that it is the center of mass system otherwise it is assumed that I have it is in the lab system and I write it by this letter L. So, the 4 momentum P A that is the 0th component we know that is E A by C again all quantities in the center of mass system and then the 3 vector that is the P A and then for the capital P B that is the 4 momentum for the particle B that is E B by C and then it is got a 3 momentum P B ok. So, how are they related with each other remember. So, the center of mass system can be thought to be moving with a certain constant velocity V with respect to the lab system. And then if our incident direction of collision is the x axis you know. So, we just have we just written the S frame on the S prime frame we just renamed it as L and L frame that is the lab frame and the center of mass frame respectively here. So, in the lab frame notice that the particle B or the body B is at rest and the particle A is coming and coming towards it with a certain momentum velocity and certain energy here kinetic energy. Of course, for particle B here in the lab system although it does not it might recall you might recall that it has some energy which is the which is associated with the rest mass energy. And then in the center of mass frame what we have we have the particles moving with the momentum towards each other equal momentum. So, that they are opposite oppositely directed and when you add them up they cancel out ok. So, how are the things related with each other recall that the coordinates in the lab and the center of mass frame they are they are Lorentz frames here. So, they are they are related by the Lorentz transformations ok. And what about the components of the four vector of the four momentum. So, since it is also a four vector momentum we are we are talking of the components of a momentum four vector. The components will transform according to Lorentz transformation itself ok. So, it is interesting that it is very easy you can you can find out if the momentum of any particle is as you know one of them in the lab frame. The in the other frame that is in the center of mass frame which is moving with a constant relative velocity with respect to the lab frame. You can simply find it out by by the help of Lorentz transformation that is the relation between the lab and the center of mass of momentum and of course, the energy the energy being the 0th component here ok. It is simple here for example, the if p x is the momentum in the small p x that is the linear momentum along the x axis in the center of mass system and then e is just the energy in the center of mass system. Then the p x l that is the x component of the momentum in the lab system is related with the help of Lorentz transformation that is gamma times p x plus beta e by c ok. So, it is just the and then since it is we have taken the motion to be along x axis. So, the x the y and the z components are the same here ok. And what about the energy part? Remember this is the for the remember this transforms like the time like components or the c t component when we are doing the position for vector ok. So, what is the energy in the lab system? So, that is simply related with the energy in the center of mass system by if e l is the energy in the lab system e l by c that is actually equal to gamma times e by c. So, e being the energy of the center of mass system plus beta times p x p x being the x component of the three momentum in the center of mass system ok fine. So, what is this velocity then with which the center of mass is moving with respect to the lamb ok. What is this velocity? How can we estimate it with the help of known components with the help of known quantities like the mass of the particles in the lab system and in the linear momentum of the particles in the lab system. So, let us find it out and to do that what we can do is that we can simply apply the conservation of a linear momentum and the conservation of momentum and also conservation of energy and we will simply take them out from there by adding the corresponding quantities for these two particles ok. So, first let us do the conservation momentum conservation first for the momentum for the for both these particles ok. And what we find is that if we add these two quantities in the lab system and then of course, we relate them to what these quantities are in the center of mass system. So, what we see is that we add p A x that is the x component in the lab system for particle A and p B x L that is the x component of the of the of the second particle B in the lab system. By the way remember that it is actually 0 that the x component that the that the momentum of the momentum of particle B is 0. So, p B x L that is actually 0 ok. So, that is equal to gamma times p A plus p B plus beta times E A plus E B by C. Now, p A plus p B so, that is 0 in the center of mass frame and then but the energies E A plus E B how are they related let us simply call them as a total energy in the center of mass frame ok. In that case we get a rather simple looking relation which relates the the component the component the momentum of A that is p A x L in in in the lab system with the total energy in the center of mass system. And remember what is this beta beta is nothing but velocity with which the center of mass frame is moving divided by the velocity of flight. And then gamma is another kinematical quantity related with beta and then the C here is the velocity of flight ok or the speed of flight if you wish. Again from energy conservation we simply add up the energies in the lab system and then corresponding and then you equate that with the corresponding quantities with the corresponding expressions in the center of mass system. And we simply also use the also use the also use the quantity that the total momentum the linear momentum in the center of mass system is 0. Then we also get rather interesting quantity ok what is it that the total energy in the in the center of mass system and the total energy in the lab system how are they related you simply take it out from this relation. But again from the conservation of momentum we have seen that the total energy in the center of mass system can be related with the total momentum in the lab system. Now we can use that relation again to to to drive out little some other quantities here. And then from from these two equations the one which relates the total center of mass and the lab energies and then the one which relates the the total center of mass energies with the lab momentum of particle A. We can find out what the velocity with which the center of mass frame is moving with respect to the lab system. We can simply work out that it is nothing but 3 square times I mean times the times the linear momentum of the particle A divided by the total energy with which the the the particle A is moving plus the rest mass rest mass energy of particle B ok. So, you have everything with respect to so in terms of everything that you started with in terms of all lab system quantities ok. So, what about the collision process? We have not used the four vectors as yet ok. We have just used the conservation of linear momentum and the conservation of conservation of energy separately and we found out this velocity. Now what if we use the conservation of the the four momentum in in the lab and the center of mass system? So, what do we get from there ok. So, for that let us simply add up the four momentum of particle A and particle B and take its norm in the center of mass system. And remember that the sum of this four vector is also a four vector and then if you take a norm of that it is going to be that is going to be invariant whether you are in the lab system and center of mass system. Let us simply call this norm as small s and given that it is still a four vector you add two four vectors you get a you get a you get a four vector I take its norm it is Lorentz invariant. So, s here is a Lorentz invariant quantity. Then what is it? It is the norm of Pa square it is it is the norm of Pa and it is the norm of Pb. These are the four vector norms plus the four vector dot product twice the four vector dot product of Pa and Pb. Now we all know how to do these four vector dot products and calculate these norms. And for example, what is Pa square that is the four vector of A square in the center of mass system. We know that it is also a it is E A square by C square that is the energy in of particle A in the center of mass system minus the three vector norm of of particle A. And then the norm of the four vector is nothing but Ma square into C square that is the Rasmus energy square of that is the Rasmus square of particle A multiplied by C square. Similarly, you can find the Pb square that is the norm of the the the vector the four vector for particle B. And then you can also find the four vector dot product of Pa and Pb that is nothing but the product of the energies minus the three vector dot product of Pa and Pb. That is a bit of an algebra. Now when you arrange all these things put it plug it back into the expression for S. What you get is that the sum of the total center of mass energy squared divided by C square minus the minus the Pa plus Pb whole squared. Now if you recall that the total moment the total the total moment of the three moment when you add these two up in the in the center of mass system so that is 0. So, the total energy in the center of mass system which is E A plus E B is a Lorentz invariant quantity here and it is and if you put it put in you know remove the square here and you see that it is actually equal to C times root S. So, S being a Lorentz invariant quantity. So, that is a rather important relation we have got here it is that the total center of mass energy for this two body collision that we see here is actually a Lorentz invariant quantity. Okay fine. Now S again is a Lorentz invariant quantity. So, we can also find the thing in the lab system okay. So, we do the thing in the lab system we take the sum of the four vector in the lab system of particle A and particle B and take its norm okay. Then with the algebra again so we have P A L L stands for the subscript L is telling us that it is a lab system P A L square plus P B L square and then twice P A L dot P B L okay. So, that is the four vector dot product. Now if we use the norm of the four vectors being invariant in being invariant here. So, we know that the norms and we also use the algebraic expression for the four vector dot products also remember that the and in using the and in writing the four vector dot product we just remember that the momentum of the second particle here B was zero. So, we are all going to have the three momentum part here in the four momentum dot product and then the energy of the particle B in the lab system is just the rest mass energy. So, we plug in all these things into the expression for the for the invariant S what we get is a quantity which is in terms of the masses and the energies of and then the energy the initial energy of particle A in the lab system okay. Just as an aside what is the kinetic energy of particle A in the lab system it is just the you subtract the rest mass energy from E AL that is the total energy and you can you can write the the invariant quantity S in terms of the kinetic energy also okay. Also we can use a approximation when you are at rather very high energies things called the ultra relativistic cases when the momentum of is actually much much greater than MA times C that is the that is the rest mass times the velocity of light okay if it is rather very big then how does this expression for the invariant S turn out to be and in that case you know that the energy of the total energy of particle A in the lab system is nothing but P AL that is the momentum in the lab system times the times the velocity of light it is can be approximated by that because the other quantity that is the quantity which depends on the rest mass will be much less compared to this okay. We may do an example for this to check it out whether that is valid or not okay. So in that case we can simply write the that the total total invariant S in terms of just the lab system momentum for an ultra relativistic case okay. We also know that the invariant S that we have defined here that we have got here can be written in terms of the centre of mass quantities remember the total energy squared divided by C squared that is also an invariant quantity actually that was also S okay. So in this way we can relate the quantities in the centre of mass system and the energy the total energy in centre of mass system with the total momentum in the in the total three momentum in the in the in the lab system. So in a way it is a it is a it is a method for us to check if you are giving this much amount of momentum in the lab system how much of it goes to the total energy in the centre of mass system okay. Okay so as I said let us do a small example okay so here let us consider the collision of two subatomic particle let us say we consider the collision of two particles and then rest masses you know the rest mass of a proton is something like 938, 940 MeV by C squared but it is approximately let us take it as 1 GeV by C squared. Now and then we consider the collision of this in the centre of mass system with three momenta of magnitude 30 GeV by C okay this actually rather very big okay. So here you can look at this figure for example you see that we have these two protons A and B they are coming towards each other with with momenta 3 30 GeV by C. Now the question is what is the what is the linear momentum what is the momentum in the lab system which will be necessary so that you have this much amount of momentum in the centre of mass system okay okay. So we simply let us find the centre of mass energy of the protons first okay. Now it will be of course the same the centre of mass energies of the more of both these protons can be found by simply you know P A squared and C squared plus M P squared C power 4 and then you take every I take the whole thing to the power half okay. So what is that so P A squared C squared what is that that is that is 930 squared okay. Now to 900 if you add 1 that is the rest mass rest mass energies squared okay. It is nothing 900 becomes simply 901 so if you take the square root of 900 or 901 it is approximately 30 okay. So it is actually an ultra relativistic case that we are talking here. So we can neglect the rest mass compared to the momenta here okay or rather rest mass times the the velocity of light compared to the momenta here okay. Now we also now we we need to calculate what is the momenta in the in the lab system okay. So for that let us calculate the invariant Lorentz invariant quantity s we know that that is related with the total centre of mass energy squared right. So s will be 30 plus 30 here because each of these protons have total energies 30 in the centre of mass system. So s is nothing but 3600 g v squared by C squared fine. We can also relate this with we have seen how it is related with the the lab system momenta okay that is actually equal to twice the mass of the protons times the velocity of light times the linear momenta okay. Now if you plug in all these quantities you will get the linear momenta of in the lab system to be right 1800 g v by C which is huge okay which is rather huge. So you see to get a energy of 30 g v per C you need such a huge momenta in the lab system okay and this also and then this number actually justifies the approximations that we have used that of our ultra relativistic case in which the momenta was considered to be the was much much larger than the rest mass times the velocity of light here. Okay so I hope I have convinced I have you know shown you some examples of relativistic kinematics more specifically relativistic collisions and the use of the full momentum or the power of the use of full momenta in relativistic kinematics. Thank you very much.