 In our last lecture, we had described the concept of force. Of course, in classical mechanics, force is well known. We discussed how we redefine it when we come to special theory of relativity. Then we realized that the way we define force in special theory of relativity, the acceleration and force need not be in the same direction. While in classical mechanics, if I apply force in a particular direction, the acceleration is also in the same direction. But in relativity, it is not like that. It need not be like that. Then we considered two special cases. One, when we are talking of only one dimensional motion, it means the force is being applied in the same direction as the velocity of the particle. And a situation in which force is applied perpendicular to the direction of velocity. In these two cases, we found out that the force and the acceleration will turn out to be the same. I mean, it will turn out to be in the same direction. So, these are the things that we discussed last time. So, we recapitulate. We discussed the definition of force in relativity. The acceleration in general is not in the same direction as force. We considered two special cases when force and acceleration are in the same direction. And this is what I have just now told. These are the two special cases. One, when the motion is along a line and the force is applied along the direction of the velocity. And the second condition, when force is applied perpendicular to the direction of the velocity of the particle. In last lecture, we had sort of said that we should, we will find out the equivalent of what we call as kinematic equations using these definition of force in relativity. So, we will spend some time today to discuss that particular aspect. Then of course, we will talk about the force transformation. So, this is the equation that we have derived earlier. Force multiplied by time. Let us just recapitulate the nomenclatures. U naught is the initial velocity of the particle. A force is applied in the same direction as the velocity of the particle. Once the force is applied in the same direction as the velocity of the particle and the motion is only in one direction, we are not talking about the vector sign. So, we have removed the vector signs. So, U naught is the initial velocity of the particle or we can even now call the initial speed of the particle. And a constant force is applied for a time t and after time t, U t is the final velocity of the particle. This is U t here. There is a U t here. There is a U naught here. There is a U naught here. And M naught is of course, the rest mass of the particle. So, this is the equation which we had derived last time, which relates force to the final velocity of the particle and the initial velocity of the particle. Of course, it assumes that force is constant as a function of time. So, my next step would be to find out this U t as a function of time so that I can directly get one equation from which I can find out the speed of the particle. This is the way the traditional kinematic equations in classical mechanics are written. So, let us attempt to do that. So, this is what I have written. The equation can be, this equation can be rewritten in a different form to enable us to calculate the speed after a given time t when a particle is subjected to a force F in one dimension motion. Of course, it essentially means reorganization of these terms because the equation has already been obtained. It is only sort of reorganizing the terms so that by direct substitution I can find out the velocity at a particular time when a constant force is applied to the particle. So, let us try to reorganize these terms. So, this was my equation which we had just now written and explained. I define a quantity A. A is not a very standard symbol. We just force divided by M naught rest mass of the particle. So, this I am calling as A which is some sort of acceleration in that sense because it has the dimension of acceleration. So, I am writing it capital A. So, if I take this, define F by M naught as capital A, then this M naught which is appearing in both these equations can be brought here. Once we brought it here, this becomes F upon M naught and this quantity will become A. This 1 upon undue to 1 minus U naught upon U naught square upon C square, I can write as gamma U naught which is the standard nomenclature that we have always been using it. So, this quantity can be written as gamma U naught. This will become gamma U naught multiplied by U naught. This quantity I take on the right hand side. So, once I have divided by M naught, this becomes this quantity becomes A t. M naught disappears from the right hand side of these two equations. This equation I take it to this side and write in the form of gamma U naught. So, this left hand side becomes A multiplied by t plus gamma U naught U naught. So, this term remains exactly identical except for M naught which we have already taken by replacing F by A. So, on the right hand side, we just have U t upon under root 1 minus U t square by C square. My idea is to write U t in terms of everything. So, what I will do? I will square this quantity and try to solve for U t. It is a simple, very simple algebra, but just to arrive at the equation which is comparatively simplified equation to determine final velocity. Because in most of the kinematic problem that we are used to in classical mechanics, essentially you are given a force or given an acceleration and you have to find out the velocity of the particle after a particular time. So, this is what I am doing in the next transparency. This is the same equation which I had written in the last transparency. This is exactly the same. I have squared it. Once I have squared it, this becomes this square A t plus gamma U naught U naught. So, A t plus gamma U naught U naught squared. When I square the right hand side, this will become U t square and in the denominator, the under root will go away. So, it will become just 1 minus U t square by C square. This I take on the left hand side. So, this will get multiplied to this particular quantity is square. So, this is what is happening here. So, this quantity after squaring and bringing into the left hand side becomes 1 minus U t square upon C square. This quantity squared is put here becomes equal to U t square. Now, it is simple. We will try to organize the terms with U t square, collect all the terms which have U t square and those terms which do not have U t square and solve for U t square. This is what I am doing in the next transparency. So, this U t square term has just been taken on the other hand side. Let me just work it out. We had 1 minus U t square by C square and here we had this quantity A t plus gamma U naught U naught square whole square is equal to U t square. So, if I take 1 and multiply this, this becomes a constant quantity and this I retain on the left hand side. So, this becomes A t plus gamma U naught U naught square. This contains U t square. So, I take it on the other hand side. So, this becomes U t square. This term will give you 1 and because this I am changing to the right hand side. So, this negative will become plus this whole quantity divided by C square. So, this is what this equation would look like. This is what I am writing in the next transparency precisely. If you look here on the transparency, it is A t plus gamma U naught U naught square and this U t square 1 plus A t plus gamma U naught U naught square divided by C square. After that simple, I divide this quantity on both the both the sides. So, left hand side becomes this divided by this square and then I take under root. Once I do that, this is the value of U t which I get U t. Once I take the under root, this just becomes A t plus gamma U naught U naught which is on the numerator. This gets divided. So, this becomes 1 plus A t plus gamma U naught U naught square by C square and of course, because I am taking under root, so this becomes under root. So, this equation would tell me if I know what is the value of A that is F upon M naught. It means if I know what is the force and if I know what is the initial velocity U naught, then it will tell me what is going to be the speed of the particle after time t. Of course, it assumes very clearly that motion has to be only along a single line because if the motion is not along the single line, these equations will not work unlike kinematic equations where you could still take the components and write those equations. Here it is very clear because in that case force and acceleration would not be in the same direction. So, in this case one has to be careful. It is only for a one-dimensional motion that I can write this type of equation and as you can see this equation is much more complicated than the usual kinematic equation that we have been used to writing V is equal to U plus A t, things like that. Now once we have obtained this equation of velocity as a function of time, I may also like to find out the distance travelled by the particle in the same time t under the same conditions. This is an equivalent of kinematic equation that we write S is equal to U t plus half A t square which tells you that if you know the acceleration, then you know what is the distance travelled by the particle in a given time t. Of course, knowing the initial velocity U, so equivalent of that equation can be found out by integrating this particular equation and that is what I will do next to find out the distance travelled by the particle in a given time t. So, this is what I have written here. The equation gives the speed of the particle with initial speed U naught after time t under a constant force F. Of course, it is very clear. Let me repeat. I assume that the motion is in one dimension, force is applied in the same direction where the velocity of the particle exists. We can integrate the above equation further to find the distance travelled by the particle. That is what I said I am going to do next. And for that matter, let us just assume for simplicity that this is along the x direction because we are talking of only one particular frame. There is no question of transformation as of now. So, you have just a frame and you would like to know, you have a particular particle which is moving in a particular direction. Let us assume that this particular dimension direction is x direction just for simplicity. Therefore, once I try to integrate this equation, this U t can be written as dx dt because the motion is only along the x direction. Therefore, if I write this U t as dx dt, this equation becomes dx and I take dt on the right hand side. So, this whole thing into dt. After that, I can integrate it. This is what I have written in this particular transparency. 0 to x dx assuming that the particle starts at time t is equal to 0 from x is equal to 0. And of course, we assume that time starts from the time when we are trying to measure the distance and goes up to the time t. Here x is the total distance travelled by the particle. I have to integrate this particular equation. Once I integrate this equation, then I can find out what will be the value of the distance that the particle travels during this particular time t. Like before, I would like to take the indefinite form of this integral and solve. Of course, left hand side is very simple because this just gives you the value of x. Because once I integrate, I will get x and if you put the limit, this will just give you x. So, left hand side is obvious, trivial. Only the right hand side has to be integrated. Though the integral looks complicated, but you can see that this integral, as we will see, the integral can be evaluated simply. So, let us go to the next transparency where I write the indefinite form. That is what I said. Let us solve right hand side integral in indefinite form by using a substitution. So, I take a substitution and using that particular substitution, I will be able to solve this integral. So, this is the right hand side integral which I have written. I have removed the limits because I want to solve this in indefinite form just to make things easy. You could have done it the other way also if you like it. And I put the substitution, this particular quantity which is under the root that is 1 plus At plus gamma U0 U0 divided by C square. Of course, this quantity also is squared. This whole quantity here within the bracket, I write as some variable that I am just taking this variable to be p. Some variable which is here which is put inside the bracket. So, once I take this particular substitution, I have to change this variable of integration also from dt to dp. For that, I have to differentiate this particular quantity. Once I differentiate this quantity on the right hand side, I will get dp. But let me first try to differentiate this particular quantity. If I differentiate 1, I will get 1. I am sorry, I will get 0. So, that does not contribute to the differential. Once I start differentiating this, this whole square is multiplied by a constant 1 upon C square or rather divided by a constant C square and C square being constant. This C square will just remain as it is. In fact, that is the reason the C square I have taken on the other hand side. Then I have to differentiate this quantity which is on the numerator. If I differentiate that particular quantity, this is fx square or rather ft square. So, first I have to differentiate assuming ft to be one single variable. So, this becomes 2 times ft. So, this becomes 2 multiplied by this particular quantity in the bracket which is a multiplied by t plus gamma u0, u0. Then I have to differentiate this quantity in the bracket itself. Once I differentiate this quantity in the bracket, I have to differentiate. I have to find out df dt. And for that, this quantity is constant. Initial velocity is constant. Of course, then gamma u0 is also constant. Then this quantity when I differentiate, I just get a. So, this a is appearing here. So, this 2 is because of this differential and this a is because of this differential and this divided by C square I have taken on the other hand side. So, what I get? 2at plus gamma u0, u0, adt is C square dp. So, what I will do? Here, I have at plus gamma u0, u0. This particular quantity I will replace by C square dp. Of course, this multiplied by dt I will replace by C square dp divided by 2a. This is what I am doing in the next transparency. This was my original integral. This quantity I have written as p. So, this becomes under root p. And as we just discussed, this multiplied by dt can be written as C square divided by 2a multiplied by dp. This is what I have written. C square upon 2a is constant. So, I can take it out. And what remains here is the integral of 1 upon under root p which becomes integral of p to the power minus half dp. This will become n plus 1 which becomes 3 by, I am sorry, minus half. So, this becomes p to the power half must be divided by half. So, this becomes 2p to the power half. So, this is p to the power half is under root p. So, this is what I have written here in this particular transparency. C square divided by 2a multiplied by 2 root p. This 2 of course cancels out. And then what I get is C square by a under root p. So, this is rather easy integral to solve, not all that difficult. I substitute the value of p and then put back into the integral. Then I will get x as a function of p. In fact, p contains time. So, now the integral as I said becomes C square by a under root p. So, this quantity was p which we have defined which of course depends on time and depends on a which is f upon m naught. We can substitute back into the integral and we can write the solution of the full integral by putting the limits. So, if I put the limit, I had to, I mean the left hand side was very clear because it was just integral of dx. So, once you put the limit, you just get x. But here when you put the limit, you have to put the limit t and subtract minus, put t is equal to 0 and then get the answer. So, this is what I will be doing in the next transparency. Let me just write it so that becomes little clear. The quantity which was written here was C square by a 1 plus a t plus gamma u naught u naught square divided by C square. So, first quantity when I put the limit t, this quantity remains identical. So, when I put 0, then I have to subtract whatever I get by substituting 0. If I substitute 0, you will get C square by a under root 1 plus gamma u naught u naught, sorry there is no square here, gamma naught u naught upon C whole square has to be subtracted from it. So, this is what I am writing in the next transparency here that x is equal to C square by a, this quantity remains as it is minus C square upon a under root 1 plus gamma u naught u naught upon C whole square. So, as you can see that this equation is also much more complicated than the simple equation s is equal to u t plus half a t square which we have been using for kinematics in the traditional classical mechanics. However, this equation can give me the distance x traveled by the particle in time t, if the particle starts with a speed u naught and is under the influence of a force f and it has a rest mass m naught. So, a is equal to f upon m naught of course, assuming that the motion is purely in one dimension. So, these were the equivalent of the kinematic equations that we are used to in classical mechanics. Now, let us go to the next step which is crucial for us to arrive at the transformation of electric field and magnetic field that we are going to discuss as the last topic of this particular course. So, question is that if I know the force of a particle in a given frame, what will happen to the force as being viewed by another frame? This is what we call as a transformation. We have been talking of velocity transformation, we have been talking of momentum transformation. What it means that I know the velocity components in a given frame and I want to find out the velocity components of the same particle in a different frame. Similarly, I know the momenta or momentum components in a given frame and I want to find out the momentum components in a different frame. Now, my question is if I know what are the force components Fx, Fy, Fz, I would like to find out what are Fx, Fy, Fz in a different frame. In classical mechanics, they are expected to be same, but as we will be seeing that in relativity they do not turn out to be same. The force components also are frame dependent. Remember, F is equal to dP dt. So, once we are talking of transformation of momentum, let us look how we are going to transform the force. So, this is what I have written as transformation of force. With the new definition of force in place, let us try to see its transformation as we change the frame. I know the components of force in a given direction. What are the components of the force in a different frame? For that, what we will do? We will go back to the concept of four vector which we have been using earlier. We had defined the Lorentz transformation in terms of four vector. We have defined velocity transformation in terms of four vector. Then we define momentum energy transformation in terms of four vector. So, let us try to construct a force four vector and if I am able to construct a force four vector, obviously that will give me the transformation of force also. So, that is the way I would look into it. I will first try to define the force four vector. As we have said that earlier we had defined the momentum four vector as this. The notation that we have used, if you remember, is that these are the four components of the four vector. First three components are the x, y and z component of momentum of the particle. First component is px, second component is py, third component is pz and the fourth component of momentum, we normally call momentum or momentum energy four vector is dependent on the total relativistic energy of the particle and in the notation that we have used we write this as IE by C where E is the total energy, total relativistic energy of the particle. Obviously, if I have to construct from momentum to force, I have to differentiate with respect to time or I have to differentiate something, I have to differentiate this with respect to something which has a dimension of time. If I differentiate this with respect to dt, again we land into the same problem when we are trying to construct from a displacement four vector to a velocity four vector because dt is a frame dependent quantity. Time interval, if you take a particular time, the moment of the particle and it little later time, the moment of the particle take the difference, this delta t time is a frame dependent quantity, that is what we have discussed. So, if I differentiate this particular thing with dt, then I will not be able to get a four vector because in that case, dt will be a frame dependent quantity. If I have to construct another four vector out of this, I have to multiply or divide by something which is a four scalar which does not depend on the frame. This we have discussed earlier and if you realize, if you remember, it is the proper time interval which is a frame independent quantity. So, therefore, if I have to construct a force four vector from a momentum four vector, I cannot differentiate with respect to dt which is different in different frame. I must differentiate with respect to dtow, the proper time interval because dtow is a frame independent quantity. If you change your frame, dtow will not change. Therefore, if I have to take a force four vector, I prefer to differentiate this with respect to dtow. That is what I have written to generate force four vector. We have to differentiate with proper time which is a four scalar. Therefore, I define force vector four vector as dp d tau where tau is a frame, d tau is a frame independent quantity. If you remember whatever we have done at the time of velocity transformation, eventually I will write to write in terms of the force component in my own frame. If I am in a particular frame and in my frame, I would prefer to write this in terms of dt as measured in my own frame. While my definition gives the differential in terms of d tau, I can always convert this d tau into dt. So, if I know the velocity of the particle at that instant of time, then I can calculate gamma u, then I know that d tau is equal to dt divided by gamma u. This we have used earlier also in the case of velocity four vector. So, this d tau can be written as dt divided by gamma u. So, this quantity force can be written as gamma u dp dt. The advantage is that now I write this thing in a particular frame. If you are in a particular frame, you would like to know things in your own frame to construct a force four vector. So, it is better that I know things in terms of my own frame. So, my prescription is that I have to calculate the momentum four vector in my frame, then take a differentiate with respect to time in my own frame, but because this is going to be a frame dependent quantity. So, what I have to do, I have to multiply this by gamma u, where gamma u is 1 upon under root power minus u square by c square, where u is the speed of the particle measured in my own frame. So, if I know the speed of the particle, I know the momentum four vector, I know gamma u and using that particular dt, I can find out what is the force four vector. So, this is what I have said. The four components of the force four vector defined in this way would then give me like this. So, I take, I calculate in my own frame what is x component of the momentum, differentiate with respect to time as measured in my frame. This is this quantity. Take the y component of the momentum, differentiate with respect to time as measured in my own frame. This will give me the second component. I take the z component of the momentum, differentiate with respect to time in my own frame. I will calculate, get the third component. I know the energy of the particle in my frame. I differentiate with respect to time in my own frame. All these quantities are given in my own frame. Then I know what is the speed of the particle. I calculate gamma u, which is 1 upon under root 1 minus u square by c square. I know gamma u. I multiply this gamma u by these quantities. This four vector is now written in respect to everything that is known in my own frame. So, I know how to write it. So, the force four vector will be given by this particular quantity. All these quantities multiplied by gamma u will generate the force four vector. If you remember, this is exactly the way we had calculated or we have constructed a velocity four vector. Now, once we know that this is a velocity four vector, I am sorry force four vector, then I know how it will transform once I go from one frame to another frame, because this is the way force transformation, four vector transformation takes place. There is a standard equation. Let me just first read this thing. The first three quantities in the bracket are equal to the components of the force. Of course, once I say dpx dt, we have already defined as the force. dpy dt I have defined as fy, dpz dt I have defined as fz and the last quantity depends on the rate of change of the total energy, which we can write in terms of power, which we have done earlier in the last lecture. So, before going to the transformation, let me just rewrite this thing. This components the force and dE dt. Therefore, the components of force four vector can be written as gamma u. First component was dpx dt, which by new definition is fx, dpy dt is according to new definition is fy, dpz dt according to the new definition is fz. And if you remember, we had calculated d dt of E in the last lecture and that we have found out to be equal to power f dot u. So, dE dt I have written as f dot u divided by c. So, like the energy happens to be the fourth component of energy moment of four vector, time happens to be the fourth component of the displacement four vector. For the force four vector, the force component turns out to be power. So, once we have generated everything in terms of forces, if I go to a different frame, in general fx is likely to change, fy will change, fz will change, of course u will change and of course gamma u will change, because the velocity of the particle would change once I change my frame of reference. I can write the transformation equation as given in this particular transparency. These are my four components of the force four vector in a particular frame S. These are the components that I want to find out in a frame S. This is the transformation matrix, which depends on gamma. Gamma depends on v, the relative velocity between the frames. See remember, we have three different speeds. So, we should not get confused. There is a particle which is moving in S. Its velocity is being measured by an observer sitting in frame S. That is what I am calling as u. Same speed, the same particle is being observed by an observer in S and that observer finds the speed of the same particle to be u. So, that means second speed. Third speed is that the relative velocity between the force, between the two frames, which is always assumed to be along the x direction and all those things, which are the conditions of Lorentz transformation. That is v. So, we have three speeds and correspondingly we have three gammas. We have gamma u. We have gamma, which depends on v, gamma u, which depends on u. This is my transformation matrix, which if I have to open up to find out what will be fx relating to other components in S frame. That is what let us do it next. So, let us expand it. I have written this particular transformation matrix here just to save some time to make it clear. If I have to write gamma u prime fx prime, this will be gamma multiplied by this component, 0 multiplied by this component plus 0 multiplied by this component plus i beta gamma multiplied by this component. So, that will be the first term. So, let us just see first term. Once I multiply it here, gamma and gamma u will come out and there will be fx. There will be gamma and gamma u, which will also be similar. i square will make it minus 1. So, second term will be f dot u by c to beta. This is what I have written here as the first term. Gamma u prime fx prime, which is the first term on the matrix, column matrix on the left hand side is equal to gamma gamma u, which we have agreed will come out, multiplied by fx minus beta f dot u divided by c. Let us look at the second component. Second equation here on the paper is gamma u prime f i prime is equal to 0 multiplied by this plus 1 multiplied by this plus 0 multiplied by this plus 0 multiplied by this, which means gamma u prime f y prime is equal to gamma u f y. Gamma u prime f z prime similarly will be gamma u f z. And the fourth equation will be i gamma u prime, this quantity minus i beta gamma multiplied by gamma u fx plus 0 multiplied by this plus 0 multiplied by this plus gamma multiplied by this. Here also you will see that there is a gamma and gamma u. There is a gamma and gamma u, which will come out, which will take out of the bracket. So, these are gamma equations. Gamma u prime f y prime is equal to gamma u f y. Gamma u prime f z prime is equal to gamma u f z. And this is the fourth equation gamma u prime i f prime dot u prime by c, which was the fourth term in the column matrix on the left hand side. And this i beta, this gamma gamma u and i comes out and you will get minus beta fx plus f dot u by c. So, these are the expanded, this particular matrix equation has been expanded into the four equations. Now, we will try to simplify these equations by using a notation or using a equation, which we had obtained at the time of velocity transformation to write in a compact form, which is a useful form and which probably is much more useful as far as our, the future endeavors are concerned. So, this is what I have written. We shall use a previous result to simplify the force transformation equation. This result was obtained while writing the velocity four vector. This was my velocity four vector transformation. So, remember this transformation matrix is exactly same. In that case, if you remember, what was forming the four vector was gamma u ux, gamma u uy, gamma u uz and gamma u ic. And when we change the frame, it became gamma u prime ux prime, gamma u prime uy prime, gamma u prime uz prime, gamma u prime ic. Let us look at only the fourth equation, which is this equation, which is gamma u prime ic is equal to gamma multiplied by gamma u, I am sorry, minus i beta gamma multiplied by gamma u ux plus gamma multiplied by gamma u ic. I am looking only at the fourth equation. First three equations will give me standard velocity transformation. So, let us look just at this particular fourth equation, which is here. Gamma u prime ic is equal to this particular quantity. This we have done earlier, discussed earlier in great detail when we are discussing the velocity transformation. And by reorganizing the term on which I would not like to put my time here, this is what we get, gamma u prime is equal to gamma gamma u1 minus v ux upon c square. So, on the left hand side of the force transformation equations, the four equations that we had got, wherever gamma u prime comes, I will replace it by gamma gamma u1 minus v ux upon c square. Therefore, I will be able to get a clean equation writing fx prime in terms of other components. This is what I am writing by the next transparency. So, I have said, we use this equation in the expansion of force transformation equation. This was my first term, which we have obtained by expanding that particular matrix equation. Gamma u prime fx prime is equal to gamma gamma u fx minus beta f dot u by c. Just now, we have obtained by expanding that particular matrix. This is the equation, which I have written in the earlier transparency, which we had obtained from the velocity transformation, which relates gamma u prime in terms of gamma gamma u. So, gamma u prime is equal to gamma gamma u multiplied by 1 minus v ux upon c square. So, this gamma u prime here, I will substitute by this particular quantity gamma gamma u1 minus v ux upon c square. Once I do here, put it here, gamma gamma u will cancel from here. What I will be getting is 1 minus v ux upon c square, which will get multiplied to fx prime. And right hand side will just become fx minus beta f dot u upon c. Now, it is rather simple. I can just divide this quantity on both the sides and I will get fx prime is equal to a quantity, which I have been doing in the next transparency. I have divided 1 minus v ux upon c square. And remember, there was a beta and there was a c here. Just here, there was a beta, beta was v upon c and there was also a c here. So, this becomes v upon c square. So, this becomes v upon c square f dot u divided by 1 minus v ux upon c square. Maybe you will see the resemblance that we get from the equation, which was a time, when we are doing Lorentz transformation, when we are doing the time transformation, we are getting something like this. It is somewhat similar to that except that the fourth term here turns out to be f dot u. This will be the transformation of the x component of the force. Let us look at the transformation of the y component of the force. Of course, let me first, before we go, let me first discuss it, because this equation sometimes is written in a different form. So, let us first work out that particular form. Then we will do the y component transformation and the z component transformation. Just wanted to mention one more point that remember fx prime will also depend on f, the total force, which means that it will also depend on fy and fz. See, like in the case of momentum transformation, px prime also dependent on the energy, which depends on p square, which will depend also on the x, y and z component of momentum, not just on the x component of the momentum. So, similarly here, it will also depend on fy and fz. So, sometime this equation is written in a different form, avoiding to write this f dot u. So, let us just try to write that equation that form before we go to the transformation of the y component and the z component. So, what we do? We just expand this f dot u. So, fx is V upon c square remains like this. And like the typical dot product of two vectors, if you have dot product of a and b, you have ax bx plus a y b y plus a z b z. So, here you have f dot u. So, you will get fx ux plus f y u y plus f z u z. This is what I have written, minus V upon c square, f dot u is fx ux plus f y u y plus f z u z. Just simply expanding and writing in this particular form. What I will do? I will take this fx common here. If I take fx common, I will get 1 minus V divided by u square, I am sorry, I will get fx 1 minus V ux by c square. So, there is a ux here, there is a V here. So, I will get V ux by c square. And that particular quantity will cancel with this particular quantity. Only as far as the x components are concerned. As far as these components are concerned, there is no cancellation. So, I can rewrite slightly this equation. This was my equation. So, as I said, fx 1 minus V upon c square, fx 1 minus V ux upon c square will cancel with this particular thing. And you will just get simple fx. These two quantities, there is no cancellation. So, you will get fx minus V upon c square, which is here, f y u y, which is here, plus f z u z, which is here. And of course, divided by 1 minus V ux upon c square, which is here. So, this tells you that fx prime is equal to fx minus all these quantities. Remember in classical mechanics, we expect V fx prime to be equal to fx. So, you can see very easily that when V is going to be very close to c, all these terms will cancel and just fx prime will be negligible. And this will give you fx prime is equal to fx. Now, let us go to the transformation of the y component. Of course, second and third terms both are similar. So, let us look first at the second term. This was the equation gamma u prime f y prime is equal to gamma u f y. This was my second equation, which we have got obtained from opening of the matrix. I use again gamma u prime is equal to gamma gamma u 1 minus V ux upon c square, exactly similar way. Here, there will be gamma. This gamma u will cancel from this gamma u on the right hand side multiplied by 1 minus V ux upon c square. So, this gamma will not cancel here. You just have f y prime into gamma into whole of this quantity 1 minus V ux upon c square divided by is equal to y f y. Therefore, f y prime is equal to f y divided by gamma 1 minus V ux upon c square. If you remember this was similar to the velocity transformation of the y component and because the z component equations are exactly identical. Therefore, I get f z prime is equal to f z gamma 1 minus V ux upon c square. Remember, in the s component, there is no gamma because that gamma has cancelled out. Here, gamma has not cancelled out, gamma u has cancelled out. So, gamma is retained in the denominator, just like the velocity transformation that we had got to obtain earlier. You can see that in non-relativistic limit, f x prime is equal to f x that we have discussed. You can also see that in non-relativistic limit, this will become 1. When V and ux are much smaller than c, this quantity is negligible in comparison to 1. f y prime will be equal to f y, f z prime will be equal to f z. So, in non-relativistic limit, we do get f x prime is equal to f x, f y prime is equal to f y, f z prime is equal to f z, but in relativity, forces also transform. In a different frame, they appear to be different. Now, let us look at the fourth component, which is component related to the power. So, power also transforms. So, once I go from one particular frame to another frame, the power will also transform. So, let us now look at the transformation of the fourth component of the fourth four vector, which is related to the transformation of power. This was my equation, which we have obtained by expanding the matrix, which was the fourth component equation. Again, I write for gamma u prime, gamma gamma u 1 minus V ux divided by c square. Here, there is a gamma gamma u, which will cancel. The i's will also cancel, which leads me to somewhat simpler equation, which is f prime dot u prime is equal to f dot u minus V fx divided by 1 minus V ux upon c square. So, this gives me the transformation of power. So, once I go from frame s to s prime, how my power changes? This is the transformation equation that governs that particular transformation equation. Now, let us look at the inverse transformation. This is tradition that whenever we write a transformation, we also write an inverse transformation. It means if my quantities are given in s prime frame of reference and I have to find out the equivalent quantities in s frame of reference, how do I transform? Like, for example, in velocity inverse transformation, my velocity components are given in s prime frame of reference and I want to find those velocity components in s frame of reference. So, that is what we call as inverse transformation. The prescriptions are simple as we have discussed the more of times that all prime quantities have to be changed to unprimed quantities and vice versa and wherever there is a V, replace it by minus V. It is as simple as that. So, that is what we are doing. We are writing now the inverse transformation equations just by changing the prime, I mean interchanging the prime and n prime and replacing V by minus V. So, this is my force component. So, on the left hand side, we have fx is equal to now fx prime. Here there was fx. So, this becomes fx prime. There was a minus V. So, this becomes because we have replaced V by minus V. So, this becomes plus. This becomes V upon c square f dot u, rather f prime dot u prime should be u prime divided by 1 plus V ux prime by c square. Similarly, f y will now become f y prime gamma 1 plus V ux upon ux prime by c square because V has been replaced by minus V. Similarly, here fz will be equal to fz prime divided by gamma 1 plus V ux prime by c square. So, this becomes my inverse transformation for force. Similarly, we can write the inverse power transformation also, exactly identical things. Here it becomes f dot, I am sorry f prime dot u prime plus V fx prime divided by 1 plus V ux prime by c square. On the left hand side, you have f dot u. So, prescription is simple. We do not have to spend too much of time describing this. Just as I said, interchange all prime quantities, make them unprime. All unprime quantities make them prime prime replace V by minus V. This gives me the inverse transformation. Before we close the lecture, let us look at a special case which is quite useful and illustrative and we will also be using in our next lecture. We consider a special case of a force transformation. Let us assume that there is a frame S in which this particular particle is instantaneously at rest. Of course, it is rest at that given time, at a later time because it is under the influence of force, so speed will change. So, it is no longer will be at rest in that particular frame. Assume that there is a frame in which at given instant in which I am interested in, this particular particle is at rest. So, whatever was the velocity of this particular particle at that particular frame, that particular frame seems to be moving in the same velocity. Therefore, this particular particle is at rest in that particular frame. In that case, u would be 0, vector u will be 0 and the force equations becomes somewhat simple equations. So, this is fx prime. This is the transformation of the x component because this particle is instantaneously at rest in S frame. So, u becomes 0. So, once u becomes 0, this whole quantity becomes 0. Of course, if u is 0, ux is also 0. So, this quantity becomes 0 and denominator we have only one which gives me fx prime is equal to fx. So, if we measure the force of the on the particle in S frame and that time when the particle was instantaneously at rest in S frame also it will appear to be at rest. I mean, it will have the x component of the force will turn out to be same. Let us look at f y prime. f y prime is equal to f y divided by gamma 1 minus v ux upon c square. ux being 0 in S frame, this quantity will be 0 and you will have just f y prime is equal to f y divided by gamma. Gamma being always greater than 1, your f y prime will be smaller than f y. And of course, because these transformations equations are similar for the z component, I will also get f z prime is equal to f z divided by gamma. So, what I get in this particular case that f x is equal to f x prime is equal to f x, f y prime is equal to f y by gamma and f z prime is equal to f z by gamma. We just see that the force is smaller because f y components have become smaller in a frame in which the particle is not at rest. So, the force is large in the frame in which the particle is at rest which you can call sort of a proper frame. So, to end the lecture, we just summarize whatever we have done. We derived the equations similar to the kinematic equations in classical mechanics. We also defined force 4 vector and discussed the transformation of force. Thank you.