 Let us start with the recapitulation of what we did in our last lecture. We discussed some simple examples involving time like and space like separated events. And just discussed how in a particular condition it is possible to find a frame in which the two events occur at the same time, but not at the same position or at the same position, but not at the same time. After that, we again took a simple example of a completely inelastic collision. These type of collisions are very well known in the classical mechanics and showed that if Lorentz transformation was the only requirement or only modification that we needed on the basis of the Einstein's postulates, then if momentum is conserved in a frame of reference, then it need not be conserved in another frame. So, either we say that conservation of momentum is not a fundamental principle of physics which I am sure many of us will object or we need to redefine a momentum. So, now we will start looking for a particular way in which I can redefine. There are various ways in which one can get an idea of how to redefine the momentum, but what I will do, I will use the method of four vectors to evolve that particular concept. The idea of four vectors, you know the way we introduce it is that we will be seeing that the universality of the conservation of momentum would be assured if we start following the way in which we work out the four vectors concept. So, we shall as take a new definition by using the concept of four vectors. But before we do that, let us first look at the concept of normal traditional standard vectors which we have been learning right from our high school telling that there are certain quantities which are scalars, there are certain quantities which are vectors. So, let me first try to re-look this particular aspect again and in a particular way so that I can lead this to the concept of four vectors. Just to mention that the terminology etc that I have followed for this particular concept of four vectors is what has been given in this particular book The Introduction of Mechanics by Klippner and Cohen. So, as I said let us try to re-look at the concept of traditional vectors. If we are somebody in high school, what is a scalar and what is a vector? They would normally say that a vector has something which has both a magnitude and a direction. So, it can be represented by something like an arrow. Of course, then many other things has to be associated with that, but that is a simple picture that every young student has in mind while there are certain quantities which do not have a direction associated with it. For example, mass of a body, there is no direction associated with that, but on the other hand weight has something in direction because weight is in terms of a force and that has a direction associated with this. So, normally we say that a vector can be represented by let us say an arrow, which has the length of which represents the magnitude of that particular vector and the way this particular error points that represents the direction of this particular vector. So, what I have done in this particular picture? I have drawn a vector which I am calling as a vector A. I have taken a particular set of axes which I am showing by this red part here. This is the x axis, this is the y axis and this is the z axis. Now, I take another set of axes. Let me remind you that I am not talking of relativity. So, the concept of S and S that we had in relativity is different. That will bring it back later, but let us take as a totally independent example. So, we have another set of vectors in which z axis is same. So, these have x prime, y prime and z prime axes while z prime and z coincides. The origin of the two sets also coincides. The only difference is that x prime is rotated with respect to x axis by an angle of phi. Similarly, y prime axis is rotated with respect to y by obviously the same angle phi. So, the only difference here is that this x prime and x they make an angle phi. Similarly, y and y prime make an angle phi. As I said again, z and z prime axes are coincident. Now, what I would like to say, what I would like to represent my vectors? If I have this set of x, y and z axes, then I can always write my vector in the form of components. I can always write my vector A in terms of the components along x, y and z direction. So, I can write this as A x and if I take a unit vector in the i direction, I can write this as A x i plus if A y is the component along the y direction, then I will write this as A j. And if A z is the component in the z direction, then I can write this as A z k, where i, j and k are unit vectors in the directions x, y and z. This is a very standard way of writing a vector in terms of the components and the unit vectors. Of course, in this case I have used the Cartesian coordinate system in which i, j, k are fixed direction along the x, y and z axes. Now, what this particular figure represents that A 5 do not use this set of x, y and z axes, but decide to use x prime, y prime and z prime axes. Of course, as far as the z component is concerned, this will remain same. But with respect to these x prime and y prime components, these components would change because now the direction of x prime is different from x. Similarly, the direction of y prime is different from y. Therefore, these components which I have earlier written would now change. So, in principle, the same vector, the vector is same, which has the same magnitude, which points exactly in the same direction. But depending upon which set of axes I decide to write the components, the components would change. So, the same vector A, which I had written earlier could now be written as A x prime, because this component has changed. Of course, this i has also changed. So, I will write this as i prime. The y component has also changed. So, I will write A y prime, j has also changed. As far as z component is concerned, it is not changed. Similarly, k is not changed, but to be little more general, let me write this as still A z prime k prime, where I will put a condition that A z is same as A z prime and the direction of k is same as direction of k prime. So, the same vector can be written in terms of two set, A x, A y, A z and A x prime, A y prime, A z prime, which represent the components of this particular vector in the three directions that I have chosen to represent my vector. So, I have chosen certain direction to represent my vector and in those directions, if I take the component of these vectors, then I can write the vector in this particular way. Now, my question is that can I find, if I know A x, A y, A z along a set x, y, z and if I know the angle phi, can I find out A x prime, A y prime, A z prime? That is the components of the same vector along the x prime, y prime and z prime direction. We can definitely do that. It is not a very difficult problem, but anyway, let us try to work it out. So, the top thing I have written, A is equal to A x i plus A y j plus A z k, which I had written also on the paper. Similarly, A is A x prime i prime plus A y prime j prime plus A z prime k prime. Of course, we agree that A z is same as A z prime and k is same as k prime. Now, what I insist that the relationship between A x prime, A y prime, A z prime and A x, A y, A z is given by this particular equation. This is not very difficult to visualize. Let us look back into my old figure and try to see this is very, very comparatively easy to derive this particular thing. I know the component of vector A along this particular direction, which is the x direction. See, as far as z is concerned, it is the same. So, let us not bother about it. Let us just look at this particular two-dimensional picture. Let me write it here. This is one set of axes. There is another set of axes, which is rotated with respect to the first axis by an angle phi and this is also rotated with respect to this by the same angle phi. This is my x prime axis. This is my y prime axis and this was my x axis and this was my y axis. This vector, the x component was somewhere here. I have to find out my x prime component. So, if I take the projection of this particular component along the x prime direction, I will get A x cos phi. Now, this is not the only component which gives a projection here. The y component will also give a projection here. Now, if this is phi, this is 90 minus phi. Therefore, the projection of the y component on this x prime axis will be A y sin phi. So, this component will give a projection in this way. This particular component will give me a projection this way. This component will give me A x cos phi. This component will give me A y sin phi. Of course, z axis is perpendicular to this particular plane or normal to this particular plane. It does not give any component in this particular direction. Anyway, we have discussed that as far as the z component is concerned, this remains same whether I decide to take x y direction or whether I decide to take x prime y prime direction. So, as we have seen that this A x cos phi plus A y sin phi would be the actual component along the x prime direction. So, this I can write as A x prime. This is what I have written here as A x prime is equal to A x cos phi plus A y sin phi. Now, let us look at the y component. If I take the y component which is y prime component rather, if I have to find out what is the component of this vector in this particular direction, I take the component along the x direction and the y direction. Whatever is the component of this, I have to take a projection on this new y prime axis. So, whatever is the component which is A y, the projection of this will be A y cos phi. If I take A x, the projection of that will be A x sin phi. But as we can say that if I take the projection of this, this will actually project on the opposite side of it. So, the component of this particular part on this particular y prime direction would be in opposite direction of this particular component. Therefore, clearly A y prime would be given by minus A x sin phi plus A y cos. So, this is what I have written in this particular equation that A y prime is equal to minus A x sin phi plus A y cos phi. And of course, we had agreed that A z prime is equal to A z. So, what we have said that there is a vector, depending upon the set of axes that I have chosen to describe its components, the components may turn out to be different. And their relationship would depend on the angle phi and this is the relationship which I write here. This particular equation, this particular set of equations can be represented in terms of a matrix equation which I am presenting in my next transparency. So, this is the matrix equation that I am writing. If I have to open up the matrix, I will get all the three components, all the three equations which I have just now written in my previous transparency. Means A x prime is equal to this multiplied by this, this multiplied by this, plus this multiplied by this. So, this will give A x prime is equal to A x cos phi plus A y sin phi plus 0. This gives me the first equation. Second equation, A y prime is equal to this multiplied by this, plus this multiplied by this, plus this multiplied by this. So, this will become minus A x sin phi plus A y cos phi plus 0, which gives me the second equation. Third equation, this multiplied by this, plus this multiplied by this, plus this multiplied by this. So, I get A z prime is equal to A z. Therefore, this matrix equation gives me the set of all the three equations which I had written earlier. So, this single equation, single matrix equation represents how A x, A y, A z will transform once I rotate the axis by an angle phi, keeping z axis same. But one thing which I would like to mention that whatever we did, irrespective of phi that I have chosen, my A x, A y can become different, they can become A x prime, A y prime. Depending upon phi, their values will be different. But all of you would agree that the length of vector will not change. The components may change irrespective, depending upon the set of axes that I have decided to work with. But as far as the length is concerned, the length will remain same, because length is a scalar. It will not depend on or it has no components. So, irrespective of whatever set of axes I decide to describe by A x, A y, A z, but the length will remain same. And if you remember, the length is defined as the magnitude or under root of A x square plus A y square plus A z square. So, this is what I call as a scalar, that this length is a scalar, because this will not change once I change my set of axes. So, this A x square under root A x square plus A y square plus A z square must be same as under root of A x prime square plus A y prime square plus A z prime square. So, though the components change, but the length being a scalar does not change once I change or once I rotate my set of axes. This is the key point which I would like to emphasize. This is probably very obvious, very simple, but nevertheless, as I said, I want to put it in a way so that eventually, we can generalize it into the form of a four vector. Let us take a simple example, very, very simple example. It is essentially a high school example, but idea is to illustrate my point. So, let us take one fixed vector A as 5 i plus 5 j plus 5 k, some arbitrary vector which I have chosen, nothing very special about these numbers. And obviously, 5 5 5 are the components of this particular vector along x direction, y direction and z direction. And I have taken just to make number simple, the angle phi as tan inverse 3 by 4. Now, let us assume that this set of axes has been rotated about z axis the way I have described earlier by an angle phi, which is given by this particular value, which is tan inverse 3 by 4. So, obviously, the same vector A would now be represented by different component, different set of components and that set of components I can find out by the transformation equation that I had just now written. So, now, let us try to find out A x prime, A y prime, A z prime using the transformation equation that we have just now described or just now written. So, my A x prime was equal to A x cos phi plus A y sin phi. A x A y both happen to be 5 in the example which I have given. So, this becomes A x prime is equal to 5 cos phi plus 5 sin phi. Similarly, A y prime will be given by minus A x sin phi plus A y cos phi. Again, A x and A y being same as 5. So, A y prime becomes minus 5 sin phi plus 5 cos phi. We have already written that tan phi is equal to 3 by 4. So, I can very easily find out what is sin phi and cos phi and this sin phi will turn out to be equal to 3 by 5 and cos phi will turn out to be equal to 4 by 5. As you can see, sin square phi plus cos square phi would be equal to 1 because 3 square plus 4 square is equal to 5 square and sin phi divided by cos phi is 3 by 4. So, these are the values of sin and cosines of phi. I substitute these values in this particular equation. If I look at this top equation for cos phi, I substitute 4 by 5. So, here cos phi gets replaced by 4 by 5. 5 cancels. So, what remains here in this particular factor is just 4. Here you have 5 sin phi. Sin phi is 3 by 5. Again, this 5 cancels out. Here you are left with 3. So, this was 4 here plus 3 here. So, A y prime becomes equal to 7 whatever units we have described. Let us not bother about the units. Now, A y prime is equal to minus 5 sin phi. Sin phi was 3 by 5. This 5 cancels out. So, this becomes minus 3 plus 5 cos phi. I substitute 4 by 5. 5 cancels out. So, what remains here is 4. So, this is minus 3 plus 4. What is left here is plus 1. So, A y prime is equal to plus 1. So, this is what I have written. x prime is equal to 7. A y prime is equal to 1. And of course, A z prime is equal to A z, which I have not specifically written. So, it means A z prime will be equal to 5. So, the same vector, now with respect to the rotated set of axes, can be represented as 7 i prime plus 1 j prime plus 5 k primes. So, as we have seen that these numbers set of numbers have changed. But as we have insisted that though these 3 sets of numbers have changed, their length will not change. And if you know, the length is given by under root Ax square plus A y square plus A z square. So, in the first case, when I had chosen x, y and z set of axes, Ax was 5, A y was 5, A z is 5. So, the length will be 5 square plus 5 square plus 5 square. This becomes 25 plus 25 plus 25 becomes equal to under root 75. In the second case, when I was representing the components in terms of x prime, y prime, so Ax prime was 7, A y prime was 1, A z prime was 5. So, the length will be given by 7 square plus 1 square plus 5 square under root. 7 square is 49, 1 square is just 1. So, 49 plus 150 plus 25 under root 75. So, as we have seen that the components have changed, but in a way as so as to maintain the length same. In fact, my transformation has ensured that the length of the vector is not changed, which should not change because this is a vector and its length is a scalar which cannot change which set of axes you have chosen to describe your vector's components. Now, let us generalize this particular idea a little more. Let us say two vectors instead of one vector. Let us suppose we have a vector A and vector B and I choose to describe these vectors A and B in terms of components of a set of axes i, j, k just like before x, y, z which have unit vectors i, j, k. So, the two vectors A and B exactly the way we have described earlier can now be written as A x is equal to A x i plus A y j plus A z k. Now, choosing the same set of axes the vector B can be written as B x i plus B y j plus B z k very, very standard way of writing the vectors. Choose a set of axes, take the components and write in terms of unit vectors along those particular directions x. We have been doing all those things essentially from high school. Now, instead of these axes if I would have chosen to describe another set of axes which are x prime, y prime, z prime with unit vectors i prime, j prime, k prime where these x prime, y prime, z prime axes are related with respect to x, y, z axes exactly in the same way as I have described earlier. It means rotated with respect to z axes. Then the components will change and again I write the same vector A as A x prime, i prime plus A y prime, j prime plus A z prime, k prime. Similarly, B B x prime, i prime, B y prime, j prime, B z prime, k prime. Now, if I take a dot product of these two vectors, I know that dot product is also a scalar. So, though the components might have changed, once I have decided to represent the same vector with respect to the components of a different set of axes, but their length will not change. I mean the dot product would not change. I am sorry, the dot product would not change because that is a scalar. So, the transformation would ensure that dot product will not change. Hence, we must have A x B x plus A y B y plus A z B z must turn out to be equal to x prime B x prime plus A y prime B y prime plus A z prime B z prime. The transformation would ensure this probably one can work out very easily and show that this really happens. This has to be actually something which is extremely obvious that the dot product to the same vector can be represent as a set of components in a different fashion by if I choose a different set of axes, but the length remains to be a scalar which would not change. Similarly, the dot product of two vectors would not change because that happens to be a scalar quantity. Now, let us go a little further, talk about cross product. We know that the product of two vectors are defined in two different fashions. The dot product is a scalar while cross product is a vector. So, if I choose the same two vectors A B and take a cross product which is A cross B, this will turn out to be a vector quantity. Now, I can again choose to describe A and B in terms of either x, y, z or x prime y prime z prime. Correspondingly, I can write A cross B, but because A cross B happens to be a vector. So, if I take the x, y, z component of A cross B and take x prime, y prime, z prime components of A cross B, they will also follow the same relationship, same transformation equation which I have described earlier. In a nutshell, what I am trying to say that if I have a set of these numbers, if they represent a vector, once I change my x, y, z axes to x prime, y prime, z prime axes the way I have described, then they will always form, they will always follow the same transformation equation. On the other hand, if the quantity happens to be a scalar quantity, then this scalar quantity will always remain same irrespective of what set of axes I have chosen to describe my original vectors with. So, that is what I have written, the x, y and z components of this cross product will also change upon rotation of axes by the same transformation equation which were used to change the components of vector A and vector B. So, rather than holding my ear like that, I will hold it like that and define or describe vector in a different fashion. I will say a vector is a set of three numbers which represent its components along the axes of a given frame or axes of a different set. And I will say that a vector is a set of three numbers A x, A y, A z which I keep in my mind are the components of this particular vector along a given set of axes. These numbers, if I chose a different rotated set of axes would change to A x prime, A y prime and A z prime, but will always obey the transformation rule which I have described earlier. However, if we have a scalar, this scalar would not change upon the rotation of axes, this will remain same. Now, I think we had enough introduction of the traditional vectors. Now, let us try to extend this idea to a four dimensional space which is generally called Minkowski space. And in this particular space, a vector is described by not three components, but four components. That is why it is called a four vector. So, as we have said, a traditional vector can be described in terms of three components. A Minkowski four vector is described in terms of four components. So, let us assume that we have one particular four vector and its components are four which is A 1, A 2, A 3, A 4. So, like when we say vector is a set of three numbers A x, A y, A z, a four vector, here is a set of four numbers A 1, A 2, A 3, A 4. We are talking of something which is little more abstract. And these components are measured in a given frame of reference S. Now, I come back to my relativity and S and S are described exactly the same way we have described earlier in Lorentz transformation. So, we still have the same S and S as we have described now in Lorentz transformation. I come back to that situation. So, A 1, A 2, A 3, A 4 are four components of something other certain four variables which have been measured in S. Then a set of these numbers four numbers would be called a four vector provided they satisfy certain transformation equation. Just take the parallel with what we have done in the case of vectors. I described in a vector three numbers which obey a certain transformation equation when I rotate the axis. Now, I say a four vector is a set of four numbers which transform when I go from a frame S to S as described in Lorentz transformation. It means there is a relative velocity between S and S and all those things. Then, of course, these numbers change, but they will change in a way which is just described by a particular transformation matrix. So, we have a set of four numbers, four variables A 1, A 2, A 3, A 4 which are measured in an inertial frame S. When these variables are measured in a different frame S, their values in general change. Like in the case of traditional vector when I rotated the set of axes, the components changed. Similarly, if I choose to describe my vector in a different frame S, these quantities A 1, A 2, A 3, A 4 would change. So, the same vector is now represented by different numbers A 1 prime, A 2 prime, A 3 prime, A 4 prime which are now measured in S frame. Earlier, they were being measured in S frame. This set of numbers, either you want to call A 1, A 2, A 3, A 4 or A 1 prime, A 2 prime, A 3 prime, A 4 prime would be called the components of a four vector if A 1 prime, A 2 prime, A 3 prime, A 4 prime bears the following relationship with A 1, A 2, A 3, A 4. And this relationship, like I have described in the case of traditional vectors by A 3 by 3 matrix, now I will describe this in terms of 4 by 4 matrix which is given in the next transparency. So, this is my transformation equation. I have set of four variables A 1, A 2, A 3, A 4 which are measured in frame S. I measure the values of the same quantities in a different frame S prime. I get the values A 1 prime, A 2 prime, A 3 prime, A 4 prime. If these four numbers are related to these four numbers or other variables given by this particular matrix 4 by 4 matrix, then I will call these four sets as forming a four vector or a Minkowski vector. Of course, I can open it, we will open it just now to when I go a little bit ahead, but this is what I want to do a third is the definition of a four vector which just is a set of four numbers like a traditional vector can be described as a set of three numbers. Here we are talking of a set of four numbers. Now, like I had described the dot product of a three vector, I can also define a dot product of four vectors. So, now consider two four vectors. In fact, this is not a very standard way of writing it, but we have just written that below A there is a curl sort of thing, sort of a wavy thing just to describe that this is a four vector. So, this four vector A has four components A 1, A 2, A 3, A 4. Take another four vector which has also a component B 1, B 2, B 3, B 4. Both are four vectors. It means both obey exactly the same transformation equation if I choose to describe these components in a different frame S prime. Now, I define the dot product of these two vectors exactly in the same way as we have been describing in a traditional just extending to the fourth dimension. So, dot product I will define as A dot B as A 1, B 1 plus A 2, B 2 plus A 3, B 3 plus A 4, B 4. Remember in case of traditional vectors we had just three components and it was A x, B x plus A y, B y plus A z, B z. Here we have four components. We write this as A 1, B 1 plus A 2, B 2 plus A 3, B 3 plus A 4, B 4. This is the way I define the dot product of two four vectors. And what I want to assert is that the transformation equation that I have described earlier for the four vector would ensure that this dot product would not change if I change from S to S prime frame. Like the dot product of two traditional vectors did not change upon the rotation of the axis. That number turned out to be same. I would now like to show that this dot product remains same even if I change my frame of reference from S to S prime provided these vectors transform the way I have written my equation. So, I have written that upon changing the frame to S prime the components of these four vectors would change as per the transformation matrix. And now you will have a different set of four numbers. So, same four vector A would not be described as A 1 prime, A 2 prime, A 3 prime, A 4 prime. Exactly the way when I rotated the axis the component of the vectors became different. Similarly, four vector B would now be described in terms of four different numbers B 1 prime, B 2 prime, B 3 prime, B 4 prime. But now we shall show that dot product of these four vectors would not change. It means I will definitely get A 1, B 1 plus A 2, B 2 plus A 3, B 3 plus A 4, B 4 equal to A 1 prime, B 1 prime plus A 2 prime, B 2 prime plus A 3 prime, B 3 prime plus A 4 prime, B 4 prime because this happens to be what we call as a four scalar. And a scalar quantity even though components may change but that particular scalar quantity their magnitude would not change. That is what is the crux of the idea. See after all this looks little bit more abstract but let us realize you know why we have why we at all evolve the concept of vectors to make our life simple. So, when you introduce the concept of vector dot product, cross product initially they all look surprising. But then we realize how much physics had become simple. See for example, if I have to define a force a Lorentz force I just write V cross B. Once I know what is my cross product I know magnitude, direction, everything I do not have to describe anything more. So, it is much easier to describe my physics with respect to standard cross product. Similarly, it becomes much simpler to describe the relativity if in terms of we write things in terms of four vectors. That is the reason we have introduced this particular concept of four vectors. Now let me try to show that this particular identity is maintained that A 1, B 1 plus A 2, B 2 plus A 3, B 3 plus A 4, B 4 is actually equal to A 1 prime, B 1 prime plus A 2 prime, B 2 prime plus A 3 prime, B 3 prime plus A 4 prime, B 4 prime. This is a small amount of mathematics. Let us just bear. This was my original transformation matrix. Let me open it up. I will get four equations out of it because this is 4 by 4 matrix. First equation is A 1 prime is equal to gamma times A 1 plus 0 times A 2 plus 0 times A 3 plus i beta gamma times A 4. Of course, I must mention that this beta and gamma are same which we have described in terms of relativity. That is beta is equal to V by C and gamma is equal to 1 upon root 1 minus V square by C square. We know about. We have used this beta and gamma a number of times. A 2 prime is equal to 0 times A 1 plus 1 times A 2 plus 0 times A 3 plus 0 times A 4. Similarly, A 4 prime is equal to minus i beta gamma times A 1 plus 0 times A 2 plus 0 time A 3 plus gamma time A 4, the standard way of opening a matrix or multiplying a matrix. So, I have to multiply these matrix. Then I equate each component. So, my A 1 prime will become gamma A 1 plus i beta gamma times A 4. A 2 prime will become equal to A 2. A 3 prime will become equal to A 3. A 4 prime is equal to minus i beta gamma A 1 plus gamma times A 4. All I have done is taken this gamma out of this. So, this gamma can be written as A 1 plus i beta A 4 just taken this gamma out of it. Similarly, I take this gamma out. So, this becomes gamma in bracket minus i beta A 1 plus A 4. So, this is the way A 1 prime would be obtained. This is the way A 4 prime would be obtained if I know A 1 and A 4. Of course, A 2 and A 3 are simple because they just happen to be equal to A 2 prime and A 3 prime. Now, let us substitute these values in the dot product that I have defined earlier. So, I start with A dot A prime dot B prime and this by the definition is written by this quantity. So, what I do? I substitute for A 1 prime and B 1 prime write this in terms of A 1 and B 1. We have just now seen that A 1 prime is gamma times A 1 plus i beta A 4. Similarly, B 1 prime will be gamma times B 1 plus i beta times B 4. The 2 gamma gets multiplied and you get gamma square. What is remaining here is A 1 plus i beta A 4 bracket multiplied by B 1 plus i beta times B 4. About A 2 prime same as A 2, B 2 prime same as B 2. So, this remains just A 2 B 2. Similarly, A 3 prime is just A 3, B 3 prime is just B 3. So, this remains A 3 B 3. Let us look at A 4 prime. A 4 prime earlier we had written as gamma times minus i beta A 1 plus A 4. Similarly, B 4 prime will be written as minus i beta times B 1 plus B 4 because there are 2 gammas multiplied. I get gamma square. So, this is the way I have written. After that the steps are simple. I just multiply these 2. I just multiply these 2 and show that this will turn out to be equal to A dot B. If I just open it here, this is what I will get. Let me try to write it here. We had gamma square A 1 plus i beta A 4. This was multiplied by B 1 plus i beta times B 4. We have gamma square. If I multiply this, I get A 1 B 1. Multiply this, I get i beta A 1 B 4. Multiply this by this. I get i beta A 4 B 1. Multiply these 2. i square becomes minus 1, minus beta square A 4. So, this is what I have written here. A 1 B 1 plus i beta A 1 B 4 plus i beta times A 4 B 1 minus beta square A 4 B 4. This quantity is exactly identical. Similarly, I can open the other bracket also, which is again very, very simple, exactly the same way and I can write it in this particular fashion. Now, I start collecting term. I will take A 1 B 1. There is an A 1 B 1 here. So, if I take A 1 B 1 out, this will have 1 minus beta square. I realize that these quantities i beta A 1 B 4. There is a minus i beta A 1 B 4. This will cancel with this. There is i beta A 4 B 1 and there is minus i beta A 4 B 1. This will also cancel with this. Then you are left with A 4 B 4 and there is minus beta square A 4 B 4. So, if I take A 4 B 4 common, again I will get in bracket 1 minus beta square. So, this is what I have written here. Same A dot, A prime dot B will be gamma square. This gamma square anyway was out here. So, this is gamma square A 1 B 1 multiplied by 1 minus beta square plus A 4 B 4 multiplied by 1 minus beta square. All other terms have cancelled out. Now, of course, plus you have A 2 B 2 plus A 3 B 3. All other terms involving A 1 B 4 and all those things they have all cancelled out. Now, you realize that gamma square is equal to 1 minus 1 minus beta square whole square. So, gamma square is equal to 1 upon 1 minus beta square. Standard beta was equal to V by C as we know. So, here you have gamma square for which I can write as 1 upon 1 minus beta square. So, if I come back to this particular transparency, if I write 1 minus 1 divided by 1 minus beta square, 1 minus beta square will cancel here, 1 minus beta square will cancel here and this equation will simply become A 1 B 1 plus A 2 B 2 plus A 3 B 3 plus A 4 B 4, which is nothing but the dot product A dot B. So, I have shown that using this transformation equation A 1 prime dot A prime dot B prime is equal to A dot B. So, though the components have changed, but they will always change in such a fashion so as to make the dot product same. The transformation equation assures this particular thing that the dot product is a scalar is a force scalar. So, it does not change its value. So, we just see that the transformation has ensured that the dot product of two four vectors does not change upon the change of the frame. Hence, it is a what we call a four scalar. Now, as a corollary of this, I need not take a dot product of A with B, I can take dot product of A with its own self. I can write A dot A and can define a length of a four vector like we describe the length of a traditional vector. So, if I take the length of a traditional four vector or sorry length of a four vector, this I can define as under root of A dot A, it means take dot product of this vector with its own self. And as we have seen from the definition, I can write this as under root of A 1 square plus A 2 square plus A 3 square plus A 4 square. And of course, this ensures that the length of a four vector would also not change once I change my frame of reference. So, if I go from S to S prime and if A 1, A 2, A 3, A 4 happen to be the component of a four vector, then A 1 square plus A 2 square plus A 3 square plus A 4 square under root will be will not change if I change my frame of reference. So, this becomes sort of a universal quantity that by changing the frame of reference, this will not change. We have talked so much in abstractness. Let me come closer to something which we are familiar with it is the standard Lorentz transformation. Now, I give an example of a four vector. I call, first I show that x, y, z and i, c, t where i is imaginary number under root minus 1, they will form the component of a four vector. This is what I call as a position four vector x, y, z, i, c, t. If they have to form a four vector, then if I change my frame of reference from S to S prime, the way I have described in Lorentz transformation, x will change to x prime, y will change to y prime, z will change to z prime, t will change to t prime and the new four components, new components of the four vector will be given by x prime, y prime, z prime, i, c, t prime and they must obey this transformation equation if they really are components of a four vector. Let us just try and test. So, x prime must be equal to gamma times x, these are zeros plus i beta gamma times i c t. This is why I have written gamma x plus i beta gamma times i c t, i square becomes minus 1, beta is v by c, gamma I can take out, this v upon c, c would cancel it. This you will get just as gamma x minus v t, which is the first equation of Lorentz transformation. So, I know, I really know that knowing x and t, this is the way x prime is going to transform when I change my frame from S to S prime. Of course, y prime, these are all zeros, y prime is equal to y, z prime equal to z, which are also part of the transformation, Lorentz transformation equation. Let us look at the fourth equation, i c t prime should be equal to minus i beta gamma times x plus these are zeros, gamma times i c t. So, I have written gamma times i c t, gamma I take common, I take it out, this is minus i, this is will cancel with this i, there is a c here, this c also I would like to cancel here, I cancel it out here, this c, if I cancel, this c will go away, then this beta was equal to v by c, there is no c here in the numerator, there is no c in the numerator here. So, if I have to take c out and cancel it, I have to divide it by c, this beta was equal to v by c. So, this will become v by c square, with this negative sign, there is a v remaining in the numerator, this will become minus v x by c square. i c I have already cancelled, gamma is taken out, so this becomes t. So, this equation will lead to me, t prime is equal to gamma t minus v x by c square, which is the fourth equation, the equation corresponding to the transformation of time in Lorentz transformation. So, I know that these equations are correct, because these equations have been described by Lorentz transformation. Hence, I know that x, y, z and i, c, t have to transform when I change my frame of reference by this transformation equation. Therefore, x, y, z, i, c, t are the components of four vector. So, then I will summarize whatever I have discussed. We have discussed the basic concept of four vector, as I generalized from the definition of a standard traditional three vectors. Then finally, I described the position four vector. We will go ahead later, evolving the concept of four vectors to other quantities. Thank you.