 In our last lecture, we had started with giving some equations which were equivalent of kinematic equations in the normal traditional classical mechanics in which we assumed that force is constant. We assumed that the motion is along a straight line. So, that force and acceleration are in the same direction even in relativity and then these equations could determine the velocity after a given time and the distance travelled by the particle in a given time. After that, we evolved the concept of force 4 vector. We had already defined what is a force. We defined in last lecture what is a force 4 vector and then eventually came to the transformation of forces. We realized that once we change the frames, the force on the body also changes in relativity. The force is also relative depending upon the frame from which you are looking, the force on the particle would also appear to be different. So, this is what we have mentioned here. To recapitulate, we found out some equations relating to the motion of a particle under constant force and we discussed the force 4 vector and then derived the transformation of forces. Now, from today onwards, we will come to essentially the last topic of this particular series of lectures on special theory of relativity, which is about the electric field and magnetic field transformation. If you recall, right in our first lecture, when we are trying to discuss the issues or the problems with the classical mechanics, at that time, we had mentioned about the Lorentz force. So, this is what I have written. While beginning our lectures, we pointed out an issue with the magnetic field in classical physics. We had mentioned that in classical physics, what this is to even relativistically, the force on a particle is given by if it is a charge q, if the particle is charged with a charge q, the force is given by qE plus V cross B, where V is the speed of the particle in a given frame of reference. Now, we had earlier discussed in the classical mechanics, in the traditional classical non-relativistic mechanics that acceleration is frame independent quantity. It means if you go to different frames, the acceleration in the traditional classical mechanics, non-relativistic classical mechanics will turn out to be same, but the velocity of the particle may differ and in general, they will be different. So, we had said that if this force is a velocity dependent force, velocity being dependent on different frames, the force would also appear to be different in different frames. Take for example, a frame of reference in which V is 0, it means the particle is at rest. In that particular frame of reference, the force on this particular particle because of the magnetic field would be 0. Let us assume that E is 0. So, it means there is no force, but on the other hand, in any other frame of reference, when V is non-zero in principle, the particle will experience a force which is V cross P or the charge particle won't experience a force which is V cross P. So, it looks funny. Of course, we know that force is a frame dependent quantity, but we do not expect that relationship to be obeyed when force is 0 in one frame of reference and in another frame of reference, it is a finite thing. Of course, at that time, we also mentioned that some people can always raise the question, having our traditional classical mechanics knowledge that after all V is also due to a motion of charge carriers. And if I change the frame of reference, I also expect that probably magnetic field will change. And that time, we had mentioned that actually the correct representation or correct transformation of magnetic field and for that matter, electric field comes when we discuss special theory of relativity. So, that is what we are going to discuss today. In fact, what we will be essentially telling is that once we change the frame of reference, both magnetic field and electric field will change. So, in principle, as far as relativity is concerned, there is no difference between electric field and magnetic field. Depending upon the frame of reference, the field may appear to be an electric field or may appear to be a magnetic field or a combination of the two. So, let us start our discussion on this particular aspect today and then we will give eventually probably some of the later lectures, some examples where we can discuss this particular aspect little bit in more detail. So, this is what we had had written. We realize that the magnetic force is a velocity dependent force and velocity is frame dependent quantities. I have just now discussed that this is the Lorentz force which is F is equal to E multiplied by E which is E is the electric field plus V cross B, where V is the velocity of the particle and B is the magnetic field. So, this is the force which will be experienced by a particle which has a charge Q or in this particular case I have written E. So, a charge E. So, now let us look for the transformation and for this particular transformation, we would essentially need the force transformation that we had earlier discussed in our last lecture and we will see that how the electric field and magnetic field should transform so that this particular force transformation is obeyed. So, we will take some specific examples and using those examples, we will derive the electric field and magnetic field transformations which appear to be somewhat more general. So, this is what I have written. We discussed at that time that the electric and magnetic field transformations are actually found out from relativity theory. So, that is what we are going to do today. So, this is what I say, electromagnetic field transformation. So, E m essentially represents electric field as well as magnetic field. So, as we said that in relativity, we do not differentiate strictly between electric field and magnetic field because what may appear to one frame of an observer in a frame electric field may appear to be mixture of electric field and magnetic field to some other observer. So, we shall obtain it by taking a specific case. The results obtained are general. Now, let us consider the case. Let us consider that there is a charge q which is instantaneously at rest in frame s. We can always assume that that though this particular charge could be moving or could be accelerating or whatever it is, but there is a frame of reference in which this particular particle or this particular charge is instantaneously at rest. If you remember in our last lecture, when we discussed in force transformation, also we considered this particular case. So, of course these frames are inertial frames. So, if actually the particle is accelerating, then if it is instantaneously at rest in this particular frame of reference, at a later time it may not be because the frames are actually inertial frames and all I am requiring that at instant of time when I am described in the situation, this particular particle is at rest in this particular frame, at a later time it may not be. Now, in this particular frame, I assume that there is a particle is experiencing in electric field which I am writing as E prime and it is also experiencing a magnetic field which I am writing as B prime. So, the particle is instantaneously at rest in this particular frame and it is experiencing electric field and magnetic field which are given by E prime and B prime. Let us read this again. We consider a charge Q which is instantaneously at rest in a frame S prime. However, it is experiencing in electric field E prime and magnetic field B prime. Once I know this particular aspect, I can always write what will be the force on the particle which will be given by the Lorentz force and now let me use the relativistic notation because this is the particle which is experiencing a force in this particular direction. So, in fact, the speed which will be given will be U prime because this is S prime frame of reference. Remember the symbol V we have reserved for the relative velocity between the frames and U and U primes are the speeds or the velocities of the particle in respective frames. So, let us write the force on this charge as the speed of the particle or the velocity of the particle is 0 at that instant in this particular frame. So, at that instant U cross B is 0. This was my force V cross B. Of course, here we have written U cross B. F is equal to Q E cross B prime. I am in frame S prime. So, everything is being primed. This is the force which is being experienced by the particle in S prime frame of reference. This is the electric field E prime and this is the magnetic field B prime. All I am insisting that this U prime is 0 at this particular instant of time. So, this force is 0. Therefore, the particle is experiencing only the force is due to only electric field. The magnetic field would not cause a force on this particular particle because the instantaneous velocity happens to be 0. So, the only force is due to the electric field. So, that is what I have written. F prime is equal to Q E prime. I can write this as component wise because this will help me. So, all I have to do is to write three components because electric field I can always write as E x prime i plus E y prime j plus E prime E z prime k. I will just take the component. So, I am writing F x prime which is the x component of the force which will be given by Q times x component of the electric field, y component of the force is Q times the y component of the electric field and the z component of the force is Q times the z component of the electric field because I am going to use this particular equation. So, what I have done? I have written this particular thing in this particular piece of paper so that I can keep on using it whenever I want. So, this is the equation which I have written here on this particular piece of paper. F x prime is equal to Q x prime, F y prime is equal to Q E y prime, F z prime is equal to Q E z prime. We will look at other equations later. Now, let us assume that there is another observer S in another inertial frame S and this is viewing this particular charge. So, there is another observer S which is viewing this chart which is I am writing here. Let this chart be viewed by another observer in frame S and let me assume that this S and S prime obey exactly the same conditions which we have been using right from beginning from Lorentz transformation. It means x axis is I mean the y axis is parallel to y prime axis, z axis is parallel to z prime axis. The relative motion is along the x direction and it is along the x axis and x and x prime axis are always coinciding. This is the standard thing which I have written here. As usual we shall assume that the axis of the frames are parallel and the origin of S prime moves on the x axis of S with speed v. The relative speed between the frames we have always reserved a symbol v. So, this is still I am using the same symbol. Now, when this particular observer looks at this particular particle at the same instant of time, it will assume that it will find out that the velocity of this particular particle is not 0. But because this particle is at rest at that instant of time in S frame. So, therefore, this particular particle velocity will turn out to be v which is the relative velocity between the frames. This also one can use the velocity transformation. We have done some examples earlier. This is very straightforward. We will always get if a particle is at rest in S prime, its velocity will turn out to be equal to v which is the relative velocity between the frames in S frame of reference. So, the velocity of the particle is v in S frame of reference. So, that is what I have said. According to S, the charge moves with a speed v along the x axis and of course, this is along the x axis. Now, we have said that in general I expect that the force on the particle, the electric field sorry, electric field on the particle will turn out to be different. And let us assume that the field that will be seen by this particular charge in this particular frame S will be E and B, electric field being E and the magnetic field being B. So, these fields in general will be different from the fields which have been seen by S prime observer using the arguments which we have just now given in the beginning of the lecture. So, what I can do, what is our methodology is that I can always calculate what will be the force on this particular particle in S frame if I assume what will be the value if I take the electric field to be E and magnetic field to be B. Then I know how the forces will transform from S to S because we have already derived the force transformation equation use these but the same equations and using those equations I will be able to find out the relationship between the electric field and magnetic field as seen in S and as seen in S prime. So, this is my logic, this is the way I am going to proceed. So, what I am going to do now is to calculate what will be the force assuming the electric field to be E and magnetic field to be B in this particular frame and then I know how the forces should transform using that I will calculate how the magnetic fields and the electric fields should transform. So, let us go to the next transparency. So, as I say according to S the charged particle is not at rest. Therefore, it would experience a magnetic force also not just the electric field, not just force because of the electric field, but it will also experience the force because of the magnetic field and that particular force will be given by f is equal to Q E plus V cross B because and this V of course is along the x direction and its magnitude is equal to the relative velocity between the frames. That is what I have written according to S the force on the charge shall be given by f is equal to Q E plus V cross B, here V is along the x direction. So, let us write this particular equation in a proper vector form and try to extract what will be fx, what will be fy, what will be fz. That is what I am doing in the next transparency. Here I want to be little more careful so therefore, because there is a magnetic field term also. So, therefore, I am writing in explicit form f is equal to Q times E xi plus E yj plus E zk. Then a term due to magnetic field plus Q V is having component only in x direction. Therefore, I am writing as Vi cross. This is the magnetic field which in general can be in any direction. So, I write B xi plus B yj plus B zk. Now, as far as this first term is concerned, it remains unchanged. I have just copied it here. This is exactly identical to this. But now, I can write this cross product and simplify the second term. I know that I cross i is 0. So, when I take Vi cross B xi, this will give me 0. So, that term will not contribute. Then I take cross product with the second term which is i cross j. i cross j would give me k. Therefore, Vi cross B yj would give me V multiplied by By in k. So, this is what I have written, V multiplied by By in k direction. Now, i cross k is minus j. If you remember, you have to go to ijk, ijk, i. So, if I go i to j, then I will get k. If I go to jk, I will get i. But here, I am getting k to i. Then I must get j. But here, I am going from i to k. So, it will be minus j. So, this is what I have written here. V multiplied by Bz, i cross k is minus j, which brings this minus sign and this is j. So, this is what I have written here. Now, like before, I will pick up the x component of the force, y component of the force and z component of the force. So, these will be the force which will be experienced by the particle in S frame of reference if the electric field was E and the magnetic field was B. Note, at this moment of time, I do not know what is E and B. I am only writing the force component and only thing which I know is the transformation of the forces. Same equation which I have written in the last transparency, I have copied it here and I am picking up the different components. So, I am writing the x component of the force, which will be the only i part. See, i is appearing only in this term, nowhere else i is appearing. So, I just have picked up one term which is q E xi. I am not writing because I am just writing the x component of the force. So, x component of the force is given by q E x. Let us write the y component of the force. For that, I have to pick up the terms containing j. This term contains j. This term contains j. So, they are both contributing to a force in the y direction. So, this will be E y minus V V B z will be the y component of force multiplied by q. So, F y will be q E y minus V B z. Now, let us pick up the z component of the force. So, I have to look only at the k terms. There is a k here. So, this is E z k. There is a k here. There is a plus sign here. So, the F z will be E z plus V B y force multiplied by q. So, F z will be q E z plus V B y. This is what I have written again here because I am going to use immediately after that in this particular piece of paper. F x is equal to q E x, F y is equal to q E y minus V B z and F z is equal to q E z plus V B y. Now, I know that what are the transformation equations? Remember, the particle is at rest in s frame of reference and in that particular case, our force transformation equations were somewhat simpler. So, those equations I can write now because the particle was at rest in s frame of reference. So, force F x of course does not change, will be F x prime. Force in y direction F y will become F y prime divided by gamma and force along the z direction F z will be F z prime by gamma. This is a special case of the force equations which we have written. Force transformation which I have used last time in which I have said that if the particles happens to be at rest instantaneously in a given frame, then the transformation equations become somewhat simpler and this precisely the case in this particular situation where the particle is at rest in s frame of reference. So, I can write these equations in rather simpler form and this is what I have written here and this also I have written in this particular piece of paper here where I have written F x is equal to F x prime, F y is equal to F y prime by gamma and F z is equal to F z prime by gamma. Now, let us look at this particular paper carefully. I know that my force transformation equation must obey these because this is the way we have derived the force. If we have to be consistent, these equations have to be obeyed. I do not know what is the relationship as yet between E x and E x prime and E y and E y prime and other fields, other components of the fields. But I can always use in this particular thing that F x must be equal to F x prime, F y must be equal to F y prime by gamma, F z must be equal to F z prime by gamma because they have to obey these force transformation equations. So, if I use this particular thing, I can quickly find out the relationships between the electric fields and the magnetic fields as we will be seeing in the next transparency. So, this is what I have written. This is the first equation which is Q E x must be equal to Q x prime which is because F x was equal to F x prime. The second equation was F y is equal to F y prime by gamma. This was my F y, F y prime was just Q E y prime divided by gamma. This is F z must be equal to F z prime which is Q E z prime divided by gamma. So, these are the three equations. First equations is straightforward, Q will cancel gives you E x is equal to E x prime. Let us look at the two equations and see what do they give us. This is just writing these equations in simpler form as I said E x prime is equal to E x. If you look at back equation, the Q will cancel out here. E y prime this gamma will come here on this particular equation and this will give me E y prime is equal to gamma E y minus V V z. Similarly, if you look at this particular equation, Q will cancel. This gamma I can take on this side. So, E z prime will be equal to gamma times E z plus V V y. So, these are precisely the equations that I have written in the next transparency. E x prime is equal to E x, E y prime is equal to gamma E y minus V B z and E z prime is equal to gamma E z plus V B z, V B y I am sorry. So, what do these equations give us? These equations tell us that if my force transformation equation has to be obeyed, then the x component of the electric field must remain identical between S and S prime framework for us. As far as the y component of the force is concerned, if electric field is concerned, y component of the electric field will change. It will become, I mean in the y, in the S prime component it will be given by E y prime. If E y and B z happen to be the y component of the electric field and z component of the magnetic field in S frame of reference, then the electric field in the y, the y component of the electric field in S frame of reference will be given by this particular equation. Similarly, the z component of the electric field will be given by this equation where E z is the z component of electric field in S frame and B y is the y component of the magnetic field in S frame. So, it means if I know what is E x, E y, E z and if I know what is B x, B y, B z, I will find out, I will find that these will be the electric field. So, I get these equations which will tell me electric field in a given frame. If I know the electric field and magnetic field in a different frame, what is interested to realize that this electric field does depend on the magnetic field in the other frame. Similarly, this electric field also depends not only in the electric field, but also on the magnetic field in the other frame. That is what we had meant by saying that relativity electric field and magnetic fields have, they do not have a really in that sense a separate context, because what this particular magnetic field is contributing so to say to electric field in a different frame of reference. So, electric field does depend on the magnetic field in a different frame. So, this is what I have written, the above equations give the electric field in frame S, when electric and magnetic fields are known in frame S. Of course, I can always do a inverse transformation. If I know what is the electric field, what are the electric field and magnetic field in S frame of reference, I can always find out the electric and I mean at the moment electric field in S frame and for that particular matter, all I have to do is to change prime to unprime quantities, change the sign of V in the standard prescription of going from a direct transformation to inverse transformation. This is what I have written, inverse transformation, it is given by usual prescription. So, I have written E x is equal to x prime, because that does not change. Here I have changed the sign, here E y is equal to gamma E y prime plus V V z prime. Here also I have changed the sign, E z is equal to gamma E z prime minus V V V V y prime. So, if I know electric and magnetic field in S frame of reference, I can calculate what will be the electric field in S frame. Of course, there is a missing component you might have realized in these equations. All I have done in this equation is to evaluate the electric field, but I have not given the transformation, I have not given the equations, what will happen to the magnetic field. We also realize that magnetic field will also change, this is what we have been saying. So, these equations only give me the values or the components of the electric field in a frame, if I know electric field and magnetic field in a different frame. I must also, I ought to calculate the components of the magnetic fields also in this particular frame, if I happen to know electric field and magnetic field in a different frame. So, for that I have to do little more exercise, which is not that simple as the way we have done it, but let us do it, because that is also equally important, because in principle I must be able to calculate both electric field and magnetic field in a given frame, if I know electric field and magnetic field in a different frame. That is what actually the electromagnetic field transformation would mean. So, this is what I have written magnetic field, I have to find out now the magnetic field components. To find magnetic field in S, when we know the electric and magnetic field in S, we need another set of equations that we have not yet found out. This is what I will do now to find out what are the components of the magnetic field, if I know electric and magnetic field in a different frame. Now, I look again to any special case and the results happen to be general enough. Now, I alter the situation little bit, which makes things little more difficult mathematically, little more difficult. In the earlier case, I had assumed that the particle was at rest in S frame of reference. Now, I will assume that this particular particle is actually moving along the plus y prime direction of S frame of reference. It is not at rest in S frame of reference, but it is moving with a velocity, but that velocity is directed along the y prime direction of S frame of reference. So, we now consider a charge q, which is moving with a speed u dot prime along y prime axis in a frame S. Of course, like before, I will assume that this is experiencing a electric field E prime and a magnetic field B prime in this particular frame, exactly the simple situation. The only thing is, which is different now that I am assuming that the particle is not at rest and therefore, the Lorentz force will also have a term because of the magnetic field. So, let us write that particular term. Again, we do exercise. The methodology is exactly identical. I take this particular force, write the components of the force, then I go to S frame of reference, then I evaluate what will be the velocity of the particle, which will now be little more complex. Assume what will be the electric field and magnetic field in S frame as E and B, then again write the forces, then I know what are the transformation of forces in the two frames and of course, in this case in neither of the frames, in general you will find the particle to be at rest. So, I have to use generic, most general force transformation and using that I will be able to find out magnetic field equations. So, let us do that. So, this is the force on this charge, which has been given as F prime is equal to Q E prime plus U not cross B prime. This is S frame of reference. In my earlier situation, this term was not there, but now it is there because I am assuming because that the particle is moving around the J direction. This first electric field term I have written just as a, in the component form E x prime i plus E y prime j plus E z prime k. U not prime I realize is purely in the y direction. So, I have written U not prime j cross this B prime. Again I have written in the component form B x i plus B y, I mean B x prime i plus B y prime j plus B z prime k. Then I will expand this particular thing. I will take this particular cross product. I will write all these terms just like before then take out x component of F prime, y component of F prime and the z component of F prime just like before. This first term is exactly identical of what I have written. This first equation is exactly identical of what I have written in my last transparency. Now let us try to evaluate this. If I take J cross i, this will be minus k because i cross j is k, J cross i will be minus k. So, you will have U not prime B x prime minus k. So, this is what I have written here. There is a negative sign U not prime B x prime. J cross j will give me 0. Then J cross k will be i. So, U not prime multiplied by B z prime i and there is a plus sign here because J cross k is plus i. So, plus U not prime B z prime i. Now using this equation I can resolve, I can take out the x components, y component, x component, y component and z component of the force in S frame of reference. Same equation which has been written here without any change. Let us pick up the x component. It means F x prime. So, let us look at those terms which contain i. There is E x prime which contains i. There is a U not prime plus B z prime which contains i. So, this becomes Q E x prime plus U not prime B z prime which forms the x component of the force. As far as the y component of the force is concerned, there is only one term which contains J. There is no other term which contains J. So, F y prime is just equal to Q E y prime. Pick up the z component, z component. This term contains k. This term contains k. So, z component is Q E z prime minus U not prime B x prime. So, these are the x, y and z component of the forces as seen by an observer in S frame of reference. Now I have just copied like before these equations in this particular paper because I am going to use these things. F x prime is equal to Q E x prime plus U not prime B z prime. F y prime is equal to Q E y prime. F z prime is equal to Q E z prime minus U not prime B x prime. These equations will be seen later. So, like before, this charge is being viewed by an observer in frame S satisfying the normal Lorentz transformation condition. And of course, I will assume that in this particular frame S, the field seen by the particle or seen by the charge is E, electric field is E and the magnetic field is B. And then again write the force and do the force transformation equations. According to S, the charge moves with a velocity U which now have to be obtained by inverse velocity transformation because I know the velocity in S prime frame of reference. I want to find velocity in S frame of reference. So, I must use an inverse transformation and if I use the inverse transformation equation, I can find out what will be the value of U in the vector form or other x and y, x, y and z components of U. Once I know what are the components in S frame of reference. So, let us do that first. These are the inverse transformation equations. X component U x prime plus V divided by 1 plus V U x prime by C square because the charge is moving only along the y direction in S frame of reference. Therefore, U x prime is 0 because it has only a non-zero component along the y direction. Therefore, U x prime is 0. So, denominator gives you only 1. This component is 0. So, U x is equal to V. It just picks up. The x component is the same as the rate of velocity between the frames. As far as the y component is concerned, of course like before because this term is 0. So, there is only 1 here. U y prime is non-zero and there is a gamma. So, this will be given by U naught prime divided by gamma because we have written the speed along the y direction to be U naught prime. So, U naught prime divided by gamma is what is the velocity along the y direction. U z is equal to U z prime divided by all these things and I know that U z prime is 0 because the particle is moving along only the y direction. So, this gives me 0. So, I have velocity 2 components. The x component is equal to the V. E is equal to V and the y component is U naught prime by gamma. So, my U has two terms as I have said. Now, using these two terms, I can again write the force. So, as I said according to S, the force on the charge shall be given by Q E where E is the field seen in this frame plus U which we have just now evaluated by inverse transformation cross B where B is the magnetic field as will be seen by an observer in this particular frame of the set where the velocity components have just been given, have just been obtained. So, I will now write this particular equation vector form and then eventually take x, y and z components of the forces then compare with what I have obtained as f x prime, f y prime and f z prime and I know that they must obey the force transformation equation. This rather longish transparency Q is equal to E plus U cross B. E like before I have written as E x i plus E y j plus E z k plus Q U I have written x component and y component, x component was V. So, I have written that as V i, y component was U naught prime divided by gamma. So, this I have written plus U naught prime divided by gamma j cross again B like before B x i plus B y j plus B z k. First term we just write like that exactly in the same way here. Second term let us take the cross product and write it open it up. So, let us first take V i cross B x i this will give me 0 because i cross i is 0. Then I take the second term i cross j will give me k. So, first term will be V by k. So, this is what I have written V by k. Then I take i cross k, i cross k will be equal to minus j. Therefore, I write this as V B z with a negative sign here j. So, it is minus V B z j. Now, I try to take cross product of the second term j cross i will be giving minus k. So, this is minus, there is a minus sign here U naught prime gamma B x k. j cross j will give me 0, j cross k will give me i plus i. Therefore, plus U naught prime divided by gamma multiplied by B z. I pick up the x, y and z component of this particular equation like before. Let us look at the x component. x component will have a ex term and you will have a term U naught prime divided by gamma B z which I have written here. y component will have a term Ey here. It will have a term V B z here. There is no other term which contains j term. So, we will have Ey minus V B z. So, what I have written here Ey minus V B z. Let us look at the z component. z component, there are three terms which contain k, this term, this term as well as this term. So, this will be Ez plus V By minus U naught prime gamma B x. This is what I have written here. So, the three components have been written here, which I am writing again here for the convenience here. fx is equal to q Ex plus U naught prime by gamma B z. fy is equal to q Ey minus V B z. fz is equal to q Ez plus V By minus U naught prime by gamma B x. Let us just look at this particular equation here and here. Now, when I am writing the relationship between fx and fx prime, I cannot use those simple relationship which I could use in the earlier case, because the particle happened to be dressed in one of the frame of references. I will use the generic relationship. So, let us write that. We have to use the general inverse force transformation to relate the forces in the two frames. See, if you remember this was the x component transformation of the force. The force term of the four vector was work, which was related to f dot U. So, fx was equal to fx prime plus V divided by c square f prime dot U prime. If I take the dot product, I am sorry, this will be given by fx. Now, because this is 0, this is 0, the one term will remain, which is f y prime plus U y prime. Therefore, I can write this particular term as just f prime multiplied by U y prime. And of course, denominator because U x prime is 0. So, this denominator, there is only one. So, I can straight away write only the numerator, which gives me fx prime plus V divided by c square f y prime U naught prime. So, this is the relationship between x and x prime. Relationship between f y and f y prime can also be formed from the general relationship. This term happens to be 1 because U x prime is 0. So, this gives you f y prime by gamma, f z gives f z prime by gamma, which is similar to what we have obtained earlier. Because U x prime happens to be 0. These are the equations, which I have written here. So, what we have just now seen that f x is equal to f x prime plus V by c square f y prime U naught prime, f y is equal to f y prime by gamma, f z is equal to f z prime by gamma. We have already evaluated f x prime, f x, f y prime, f y, f z prime, f z. All we have to do is to substitute in these equations. Let us first take only the x component of this equation. Substitute this value of f x here, which is q E x plus U naught prime by gamma V z here, f y prime is equal to q E x prime plus U naught prime V z here. Then there is f y prime, f y prime we put q E y prime. So, I substitute these things in this particular equation and let us see what we get. So, what I have written using the equation corresponding to the x component, we get q E x plus U naught prime by gamma V z, which happens to be f x. This what is f x prime plus V by c square, this is what was f y prime. Using this particular equation, you can see that q gets cancelled out and I can write this particular equation with E x prime is equal to E x because E x prime is equal to E x has already been evaluated, has already been found out when we did transformation obtaining the E x E y and E z components. So, E x prime we know is equal to E x. So, what I will do is cancel this q, substitute E x prime is equal to E x. So, using the following transformation that is E x prime is equal to E x, we can write the equation as follows, which I am writing here. q has been cancelled out. Instead of E x prime I have written E x, as you can see that this E x would cancel out. Here we have U naught prime divided by gamma B z. Once this E x has been cancelled out, there is a U naught prime here, there is a U naught prime here, there is a U naught prime here, this U naught prime will also get cancelled out. So, here I will only get B z by gamma. This gamma I can take on the right hand side and multiply it. So, B z will become equal to gamma, which is this gamma which gets multiplied here. This term has already cancelled out with this particular term. U naught prime has already cancelled out B z prime, which is B z prime plus V by c square, which I have written here. U naught prime has already cancelled out and this is E y prime, which I have written here. So, as you can see I get a equation B z is equal to gamma B z prime plus V by c square E y prime, which tells me the z component of magnetic field in S frame of reference. If I know the z component of the magnetic field in S frame of reference and electric field, y component of electric field in S frame of reference. So, this is one of the equations which we have obtained using these force equations, which is part of the velocity, magnetic field transformation giving me the magnetic field. So, as I have written, this gives the inverse transformation of the z component of the magnetic field. Now, when I look back at this particular paper, this particular equation F y is equal to F y prime by gamma, you can try it yourself, does not lead immediately to a new equation. I will use this third equation F z is equal to F z prime by gamma. So, for this F z, you know that there are three terms U E z plus V B y minus U naught prime by gamma B x and for F z prime, you have two terms U E z prime minus U naught prime B x prime. So, in this equation, I will substitute this value of F z. Here, I will substitute this value of F z prime. This is what I am writing in the next transparency. So, let us use the equation corresponding to the z component of the force. So, this is this gamma I have multiplied because on both the sides. So, I get gamma times Q E z plus V B y minus U naught prime by gamma B x. This quantity was equal to F z and this was F z prime, which was actually divided by gamma, gamma I have taken on this side. So, this becomes Q E z prime minus U naught prime B x prime. This E z prime, I know how to convert in terms of E z by using the transformation of electric field, which we have just now obtained. So, this is what I will be substituting in the next thing. Using the following transformation, E z prime is equal to gamma E z plus V B y, which we have just now derived. I will substitute in this equation and I will get this equation. Left hand side remains exactly identical. Q has been cancelled out. For E z prime, I have written gamma E z plus V B y and this term remains as it is. Now, if you can see very clearly, this gamma E z will cancel out with this gamma E z. There is a gamma V B y which cancels out with this gamma V B y here. If you look at this term, this gamma will cancel with this gamma. You have minus U naught prime B x. On the right hand side, you will have minus U naught prime B x prime. U naught prime and U naught prime will cancel. For these minus signs, you can always cancel. This gives me B x is equal to B x prime. So, like behind E x is equal to x prime, I get B x is equal to B x prime. So, this gives us the transformation of the x component of the magnetic field. So, now I know how to find out B z and how I know how to find out B x. I have still not been able to find any equation corresponding to y, which requires a little bit more effort. For obtaining the transformation of y component of the field, we have to assume that the charge is moving in the plus z direction, z prime direction like we have assumed in plus y prime direction in s prime frame of reference. So, we do exactly the same calculation again, but now we assume that the charge is moving in plus z prime direction in s prime frame of reference. Otherwise, we do all the calculations exactly identically. We can carry out exactly similar analysis as we have done just now. The first equation would give the transformation of y component of the magnetic field. Just like in this particular case, it had given me the transformation of the z component of the field. If I would have assumed that the charge is moving in plus z prime direction in s prime frame of reference, the first equation will give me the transformation corresponding to the y component, which we will not work out here. One can try to do it yourself and the result that we will obtain will be equal to B y is equal to gamma B y prime minus V by c square E z prime. So, now I have found out all the three inverse transformation corresponding to the magnetic field. I can always find out the direct transformation by doing the same prescription replace V by minus V prime by unprime, unprime by prime. So, these are all my equations which are relating to the magnetic field. This is what we call as the direct transformation. I know electric field and magnetic field in s frame of reference. I will find out the magnetic field in y, s prime frame of reference. Here I know the electric and magnetic field in s prime frame of reference. I can find out the magnetic field components in s frame of reference. We have already found out the relationship, which gives me E x, E y, E z, E x prime, E y prime, E z prime. Essentially means that for example, if I know what are the magnetic field and electric field in s prime frame of reference, I can always find out what will be the electric field and magnetic field in s frame of reference or vice versa using these equations. So, essentially we have three sets of equations corresponding to the electric field components and three equations set of equations corresponding to the magnetic field components. So, if I know electric field and magnetic field, I can find out electric field and magnetic field components in any other frame of reference. And as you can see very easily from these equations, they are all interdependent. For evaluation of electric field, I also need to know the magnetic field. For evaluation of magnetic field in general, I also need to know the electric field. So, here I come to the end of this particular lecture. I just summarize. In this particular lecture, we have done lot of special calculations to find out the transformation related to electric and magnetic field. Thank you.