 In our last lecture, we had described the concept of electric field and magnetic field transformation. We had said that in relativity, a field can be mixture of the two. What it means that if in a particular frame, the field experienced by a charge is purely electrical, it is likely or in general, it will always be that in any other frame, it will experience both a combination of electric and magnetic field. So, what we generally call is an electromagnetic field, because it is nothing purely electrical, purely magnetic. It depends on the frame from which you are viewing. So, this is what I am recapitulating. We discussed the transformation of electric and magnetic fields. Using this transformation, we found out that if I know the field, electric and magnetic field in a given frame, how do I find out the electric and magnetic field in a different frame? This is what we had discussed in our last lecture. Today, what I will try to do is first to tell that some of these equations can be derived in a different fashion, because they happen to be somewhat interdependent. So, that is the first aspect that I will discuss. Then I will give one example, one specific example of how electric field and magnetic fields will transform once I go from one frame to another. So, this is the electric field transformation that we had written. If I know in a frame S, the electric field components, it means I know E x, E y and E z. And I know the magnetic field components that is B x, B y and B z. Then, in a different frame S, of course, S and S frames have to have the same relationship as we have discussed in the earlier Lorentz transformation. This E x will turn out to be equal to E x, E y will turn out to be equal to gamma E y minus V B z. And E z will turn out to be gamma E z plus V B y. Similarly, we can write the inverse transformation, which means that if I know the fields in the primed frame of reference, I can find out what is what are going to be the electric fields in unprimed frame of reference or in S frame of reference. Now, these equations tell only half the story, because they only tell you how to evaluate the x, y and z component of electric fields. We also require equations which can tell us what will be the x, y and z component of the magnetic field. So, there is a different set of equations which gives the transformation or which enables us to find out the magnetic field in a frame if I know the electric field and magnetic field in a different frame. So, these are the set of equations. So, B x prime turns out to be equal to B x, B y prime turns out to be equal to gamma B y plus V by c square E z, B z prime turns out to be equal to gamma B z minus V by c square E y. Similarly, this inverse transformation, the fields are known in S frame of reference and using these equations, I can find out what will be the magnetic field in the unprimed frame of reference or in the S frame of reference. So, these are the equations which enable us to find out the magnetic field. So, this in combination with the earlier three equations, which are about electric fields, I can find out all the components, the electric field components as well as the magnetic field components in a given frame. Now, what I wanted to tell is that these last two equations which we have obtained here, for example, can also be derived using the last two of electric field equations, which are these equations. So, let me first discuss these particular aspects because there is some sort of interdependence of these equations on each other. So, let me first discuss that particular aspect before we come to an actual example. So, this is what I have written. Before we proceed ahead, you would like to demonstrate that the transformation of y and z component of magnetic fields can also be obtained from direct and inverse transformation of electric fields. So, this is what we will try to initially try to show. So, what I will do? I will first write the second equation as I said from both the sets. Here on this particular piece of paper, I have already written these equations. These are the equations which enable us to determine the electric field in a particular frame of reference. So, let me look at the second equation. This is this equation as well as this equation. This equation tells me that if I know E y and V z, I can find out E y prime. This equation tells me that if I know E y prime and V z prime, I can find out E y. So, what I will do? I will try to eliminate E y prime from this particular equation using this equation and as you will be able to see that from this we can obtain what will be the value of B z prime or the magnetic field transformation equation. Similarly, if I use these two equations and eliminate E z prime, I can find out a transformation which gives me the value of B y prime if I know the electric and magnetic fields. So, let me first attempt that. So, these two equations as I have said are the second equations which I had just now described. So, E y prime is equal to gamma E y minus V B z and E y is equal to gamma E y prime E y prime plus V B z prime. This E y in this particular equation, I will substitute this value of E y prime. So, here in place of this E y prime, I put E y prime is equal to gamma E y minus V B z. So, this is what I have put it here. This is this particular quantity which is here is replacing this E y prime. The second term term is as it is V B z prime. Now, what I will do? I will expand it. I will multiply here. So, once I multiply E y and V B z by this gamma, it has to be further multiplied by gamma, it will become gamma square. So, let us look at this particular equation. This is the equation which I have just now written on my last transparency. So, as I said, when I open it, when I expand this particular bracket, there is a gamma here. There is another gamma here. So, this will become gamma square. This will become gamma square E y. Similarly, there is a gamma here. There is a gamma here. There is a negative sign here. So, it will become minus gamma square V B z plus this term remains as it is except that this gets multiplied by gamma, which is here. So, it becomes gamma V B z prime. Now, what I am trying to do is to find out the value of V z prime. So, I want to take this on the right hand side. If I take, want to take this particular thing on the right hand side, I will get the equation which is this will become negative. So, this becomes let me first look at this. If I take this on the left hand side, I am sorry and E y on the right hand side, then this V z prime I have to divide by gamma. Let me write this particular thing to be more clear. The equation which I had was E y is equal to gamma square E y minus gamma square V B z plus gamma V B z prime. So, I take this on the left hand side or you can put I take these things on these two things on the left hand side because this I want to retain as positive quantity. So, I can write this as gamma V V z prime is equal to E y minus gamma square E y. When this goes to this side, this will become plus gamma square V B z. I divide the equation by gamma V. If I divide the equation by gamma V, I will get B z prime is equal to E y by gamma V minus gamma square E y divided by gamma V plus gamma square V B z divided by gamma V. This is the equation which I have written here. Sorry, this involved a few steps. So, B z prime is equal to E y divided by gamma V minus gamma square E y divided by gamma V plus gamma square V B z divided by gamma V. Same equation which we just now wrote on this particular piece of paper. Now, this gamma square, there is a gamma square here, there is a gamma here. So, this will cancel with one of the square and this becomes just gamma. Similarly, there is a gamma here, there is a gamma square. So, this will become gamma. Here, this V and this V will cancel out. So, this particular term, the third term just becomes gamma B z. Here, you have just gamma E y by V and this term remains as it is E y by gamma V, a simple algebra. Put this equation here. From these two, I take E y common. So, I write E y. What is remaining in the bracket is 1 divided by gamma V minus gamma by V. This term is gamma B z as it is. Now, what I do? Out of this particular equation, I multiply it further by gamma. So, this equation becomes gamma E y by V. If I multiply by gamma, these things have to be divided by gamma. So, I divide by gamma. When I divide this term by gamma, I get 1 upon gamma square. When I divide this term by gamma, I just get 1. This V has been taken out here. So, this equation becomes gamma E y by V. V comes from these two equations and 1 minus 1 divided by gamma square minus 1 plus gamma B z. Now, this equation can be simplified because I know what is the value of gamma square and therefore, I can always find out what will be the value of 1 upon gamma square, which I am doing in the next transparency. As we know, gamma square will be equal to 1 upon 1 minus V square by C square. So, if I take inverse of gamma square, when this we have used number of times earlier also, it becomes 1 minus V square by C square. So, here it becomes 1 minus V square by C square. This one will cancel with this one and what will be remaining is just minus V square by C square. So, this term remains here and this whole bracket gives me minus V square by C square. So, this is minus V square by C square plus gamma B z. This gamma can be taken out. I get B z minus V by C square E y, which as you would have realized that this is the transformation equation which gives me the value of B z prime, the z component of the magnetic field. If I know the z component of magnetic field in S frame and the y component of the electric field in S frame of reference. Similarly, you could have taken the third set of equations as I have described just now and eliminate E z prime from that particular equation and then I could have obtained transformation equation corresponding to the y component of the magnetic field. In that case, B y prime will turn out to be equal to gamma B y plus V by C square E z. So, as we have seen that the last two equations of magnetic fields transformation can also be obtained just by using only the electric field transformation and just manipulating the last set of equations. Of course, you have to use the direct transformation as well as inverse transformation and you can manipulate these equations to get this result. I would just like to mention one more point because many times, I mean of course it is always we have always been telling when we started special theory of relativity that the initial conditions that we have put on S and S prime, the relative velocity was always along the x direction. But sometimes, we will prefer to write a little more generic equation which does not have this relative velocity between the frames only along the x direction. And if you want to do that, these equations of electric field and magnetic transformation can also be write it along the longitudinal and the transverse direction. That is the direction parallel to the relative velocity motion direction and perpendicular to the relative velocity motion direction. So, this is what I have written in case we do not want the relative velocity direction between the frames along x direction, we can write the perpendicular and parallel to the relative velocity components. So, when I say perpendicular, it will mean perpendicular to the relative velocity between the frames and parallel which means parallel to the relative velocity between the frame. The results somewhat obvious, but nevertheless let me just give because this sometimes we give these set of equations. In such a case, the transformation equations for the electric field can be written as. So, many times you will see in the text book they write this type of equations that E parallel prime is equal to E parallel, E perpendicular prime is equal to gamma E perpendicular plus V cross B, the perpendicular component of this cross product. So, essentially the other two equations which we have written y and z component can be merged into this particular way of writing these equations. So, basically you will have to remember in that case only the two equations. Of course, the first equation is obvious because along the direction of the relative motion the fields do not change. It is true for electric field as well as the magnetic field because even for magnetic fields we got Vx is equal to Vx prime. It is only along the perpendicular to the relative velocity motion direction that the fields transform. So, this is what we call as a perpendicular or transverse transformation of the fields. Similarly, we can write the equation corresponding to the magnetic field also. So, for magnetic field I will write V parallel prime is equal to V parallel and V perpendicular prime is equal to gamma V perpendicular minus 1 upon c square. For this here there is a term of 1 upon c square V cross E perpendicular component of this particular thing. And if you have any worry you can always take V along the x direction and expand the cross product you will get the same three equations that we have derived earlier. So, what I have written this can be verified by taking the velocity along x direction and expanding the cross product. So, this was about the electric field and magnetic field transformation. Let me now take one example, one simple example extremely simple example just to give you some idea of how the fields transform. So, we consider three different frames of reference S, S prime and S double prime. We of course assume that their y and y prime and y double prime axes are parallel in all those things which we have always assumed for deriving the original Lorentz transformation. Now, the speed of S prime and S double prime in S frame is 0.6 c and 0.8 c respectively in plus x direction. So, what has been given are the velocities of S prime and S double prime in S frame. The speed of S prime and S double prime in S frame is 0.6 c and 0.8 c. So, we have three frames let us say this is S frame we have S prime frame and we have S double prime frame. Now, the speed of S prime in S frame has been given as 0.6 c. So, S prime in S frame appears to be moving with a speed of 0.6 c of course, along the plus x direction. Similarly, S double prime is speed has also been given only in S frame which happens to be 0.8 c. So, the speed between these two frames is 0.8 c. So, both the speeds have been given in S frame. Let us go ahead with the question. In the frame S prime magnetic field is 0. So, what has been told that in S frame the magnetic field is 0. While there is an electric field along the y prime direction, put the direction of y prime is same as the direction of y because that is the way our transformation equations have always been written. And its magnitude is E 0 prime or E naught prime. So, this is the magnitude of the electric field which is happening which is only in the plus y direction. So, a person in the frame S says that there is no magnetic field. There is only an electric field which is pointing out in plus y prime direction and it has a magnitude of E naught prime. Now, what I have to do is to find the field in S frame, find the field in S double prime frame and we have to also find out the forces between all these three frames S frame, S prime frame and S double prime frame and see verify ourselves that the forces actually obey the force transformation equation the way we have written it earlier. Of course, this is obvious because these equations we have been derived by using the force transformation. So, this has to be valid, but the idea is to take a simple example and convince ourselves because nothing like working at our own self and try to convince ourselves that whatever I am getting is consistent as I would have expected. So, this is what I am trying to do in this particular example. So, as far as the field information is concerned, it has been given in S frame of reference. We have been told that the magnetic field is 0. It essentially means that B x prime is equal to 0, B y prime is equal to 0 and B z prime is equal to 0. We have also been told that electric field is only in the plus y direction. It means there is no component of electric field in x direction and z direction. Therefore, E x prime is equal to 0 and E y prime is equal to E naught prime which has been given. It is a constant field which has been given as E naught prime. Of course, we have not mentioned about constant, but does not matter. And E z prime is equal to 0. The only field component which is existing in S frame is E prime is equal to E naught prime. All other components are 0. Now, we have said that S moves in S with speed 0.6 c. Hence, to get the fields in S, we have to apply inverse transformation with relative speed v is equal to 0.6 c. Let me just look back at this particular field which I have written earlier in this particular piece of paper. Here, we have written in this particular piece of paper that if you look at this particular frame, S and S frame of reference, S moves with a relative velocity of 0.6 c in frame S. Therefore, if any information which is given in S has to be transformed to S prime, we will use the direct transformation. If the information has been given in S prime and I want that information, equivalent information in S, then I have to use the inverse transformation. Now, here the fields have been given in S frame of reference. Therefore, I must use the inverse transformation to find out the field in S frame. The relative speed is 0.6 c in S frame. So, I still use v is equal to plus 0.6 c, but use an inverse transformation because the fields have been given in S frame of reference. So, if you remember the original Lorentz transformation, the way we have derived, this comes from that, that if things are given in S frame of reference in S frame, if I have to find out, I have to use the inverse transformation. So, this is what I will do. The component of the electric field in S, of course, you know, we have done number of times that for 0.6 c, the gamma will turn out to be equal to 1.25. So, I do not want to derive it again. Gamma is equal to 1 upon under root 1 minus p square by c square. You can, if you do not remember, you can always work it out that if I take v is equal to 0.6 c, gamma turns out to be equal to 1.25. Now, I have to just use the inverse transformation equations which I had written it earlier here. On the right hand side on this particular piece of paper are the inverse transformation equations. E x is equal to E x prime. This is what I have written. E x is equal to E x prime. We have already seen that E x prime is equal to 0. So, E x also has to be equal to 0. E y, I have written this equation. E y is equal to gamma E y prime plus v b z prime. This is what I write here. E y is equal to gamma E y prime plus v b z prime. Here only E y prime is non-zero quantity. E b z prime is 0. That is what we have said earlier. E y prime is equal to E naught. So, this becomes just gamma times E naught. Gamma we know is 1.25. So, this becomes E is equal to 1.25 E naught. Of course, there is a prime. Similarly for E z, we write this equation because this is 0. This is 0. Therefore, E z is equal to 0. So, what we have seen is that only the electric field, the y component would be present out of the three electric fields. E x will be 0. E z will also be 0, but E y would appear to be changed in principle. E y will turn out to be larger than E naught prime. It will become 1.25 times E naught prime. But that is not the complete story because what I have calculated are only the electric field. I must also calculate the magnetic field to find out if there is a magnetic field in S frame. In S prime frame, there was no magnetic field, but in S frame, there could be a magnetic field. So, the components of the magnetic field can be derived or can be found out by using these equations. Let me just remind you these equations which I have written here in this piece of paper. These are my equations. I have to use an inverse transformation. I would use this set of equations. B x is equal to B x prime as we have said along the x direction, which is the relative motion velocity direction between the frames. The fields do not change. So, B x remains equal to B x prime. B x prime was anyway equal to 0. So, B x is equal to 0. This is what I have written here. As far as second equation is concerned, B y prime is equal to gamma B y prime minus B y c square E z prime. E z prime is 0. B y prime is equal to 0. So, this is also equal to 0. So, as far as x component and the y component of magnetic fields are concerned, they are 0 in this frame also. But let us look at B z, which is non-zero. B z, if you look back in this particular paper, B z was equal to gamma B z prime plus V by c square E y prime. Same thing I have written here. B z prime is 0, but E y prime is not 0. It is actually equal to E naught prime. Gamma is equal to 1.25. V, I know is 0.6 c. So, this is 0.6. This c will cancel one of those squares. So, I will get 0.6 multiplied by E naught prime divided by c. So, if I just calculate this number, this turns out to be equal to 0.75 E naught prime by c. So, what I find that in this particular frame of reference S, the electric field was equal to 1.25 times E naught prime, which we have just now written earlier. And you have B z, which is equal to 0.75 times E naught prime divided by c. So, what I see that in S frame of reference, the charge or the person in the frame will also experience a magnetic field in the z component direction pointing out in the plus z direction, which will be equal to 0.75 times E naught prime by c. So, the field which was purely electrical in S frame of reference turns out to be partly electrical, partly magnetic in S frame of reference. So, what I have written here? We thus see that the electric field in S is 1.25 times the same in S prime. Moreover, a magnetic field would also be felt in S frame pointing out in plus z direction. This is what comes out from the relativity. Now, let us look at the forces. If I look at the force in S frame of reference, because there was only an electric field and also, it has been given that the charge is at rest, but anyway there is no magnetic field there, which has not been given. So, we can write the force, which I can write as E y. Once I write E y and because I am sorry, let me write the force, not the field. Field has been given. I am writing F y. This will be given by just E times E y, because that is the only force, because there is electric field in the y direction and there is no magnetic field. Charge is anywhere at rest. So, let me write Q, because Q is the value of charge, which has been given to me. So, the force will be just given by Q E y purely electrigraphy force, which will be pointing out in the y direction. So, this is what I have written. F prime is equal to Q E naught prime J. This I have written in the vector form. Of course, i, j, k point out in the same direction. They just give you unit vectors irrespective of the frame. They give the unit vectors in directions pointing out along the y direction, x direction, y direction and z direction and because this is j. So, therefore, they are pointing out in the y direction. Now, let us evaluate the force on this charge in S using the calculated fields. Now, as far as S frame is concerned, there is also a magnetic field and both the charge is at rest in S, prime frame of reference. Charge is not at rest in S frame of reference, because S moves relative to S. Therefore, the charge also moves relative to S. Therefore, if there is a magnetic field in S, the magnetic field will also cause a force on this particular charge. So, let us calculate that particular force. Now, we are looking at the force on the particle. So, let us first look at the speed of the charge, because the charge is at rest in S, prime frame of reference. So, whatever is the speed of S, prime in S frame of reference, same would be the speed of the charge. So, u, which is the speed of the charge, as seen in S frame of reference, will be just 0.6 c and it will be pointing out in the i direction, because S frame of reference, rather S, prime frame of reference moves in S with a speed of 0.6 c in plus x direction. So, therefore, this will be u. I can write the force. Force will be given by q times e, electric field we have already seen is 1.25 times e naught prime plus v cross b, v is 0.6 c i. We have already seen that there is a magnetic field pointing out in plus direction plus z direction and the magnitude of that particular field is 0.75 e naught prime by c and because it is in plus z direction, I have written k. So, this will be the entire Lorentz force on this particular particle. This is because of the electric field. This is because of the magnetic field. First term I keep as it is. Let me look at this cross product. This is i cross k. So, i cross k will give me minus j. I just multiply 0.6 c with 0.75. This c will cancel with this c. You will get minus 0.45. This minus sign is because i cross k is minus j. So, this gets subtracted from this and I find that the force which we experienced in charge in s frame of reference will be 0.8q e naught prime, 0.8q e naught prime j. So, this piece of paper gives me what are the field values that I see, what are the speed of this particle and what will be the force on this particular charge. Remember, this is q. Of course, I expect that this force must obey the transformation equations that we have found out. In s frame, the force is in the y direction. So, it is an s frame of reference. So, let us look only at the equation corresponding force transformation corresponding to the y component. And if you remember, if charge happens to be at rest or if the particle happens to be at rest in one of the frame, the transformation equation is comparatively simple. So, I will use those simple equations. This is very obvious here because it is in the prime frame of reference that the charge is at rest. So, force, the y component of the force will be just given by f y prime by gamma and gamma is 1.25. 1 upon 1.25 is just 0.8 and this is what I have obtained that the force has become 0.8 times the force on the charge in s frame of reference. So, this obeys the force transformation equation as I would have expected. Now, let us try to calculate the fields in s double prime frame of reference. And let us, I mean I could have gone either from s frame of reference or I could have gone from s frame of reference. So, at the moment, I am transforming from s frame because now I know the fields both in s frame as well as in s frame of reference. So, I could have used any transformation. Let me first calculate the fields starting from s frame of reference where I have calculated the fields just now. At later part, I will also transform from s frame of reference to see that we are getting everything right which we must get. We must test ourselves that whatever we are doing is real turning out to be correct. Now, if I go from s frame to s double prime frame of reference, I go back to this particular piece of paper here. This is s, this is s double prime. We have said that according to an observer in s frame of reference, s double prime frame moves with a speed of 0.8 c. So, if I have to, if I know certain quantities in s frame, I have to transform into s double prime frame of reference. Therefore, I will use direct transformation. And I will use a direct transformation with a v is equal to 0.8 c because this is the relative velocity between the frame. So, remember the symbol v was reserved for relative velocity between the frames. So, here I will use v is equal to 0.8 c because that is the relative velocity between s and s double prime. And I will take v is equal to plus 0.8 c and I will use a direct transformation because the information is known as s frame and I want them to be in s double prime frame of reference. Now, again v is equal to 0.8 c gives me a clean number for gamma and this gamma happens to be 5 by 3. This also we have done in many of our earlier examples. We have used corresponding to 0.8 c, gamma turns out to be equal to 5 by 3. Now, s double prime is equal to x. There is no problem because as I have said along the relative velocity direction, the fields do not change. So, relative velocity is along the x direction. So, the x component of the fields do not change. So, e x double prime is equal to e x is equal to 0. Now, let us look at e y double prime. Here I am going to use the direct transformation equation as I have mentioned earlier. If I look at this paper, earlier when I went from s prime to s prime, I use this equation. When I go from s to s double prime, I use these equations because now I am using a direct transformation. So, I use e y prime is equal to gamma e y minus v v z. And remember in s frame, we have nonzero e y and nonzero v z. Both are nonzero. So, I should calculate e y prime. This is what I have done here. E y double prime is equal to gamma e y minus v v z. We have just now calculated the electric field in s frame, which is 1.25 times e naught prime minus v. v is the relative velocity between the frames, which is 0.8 c. v z is the magnetic field as seen in s frame or the z component of the magnetic field, which is 0.75 times e naught prime by c. This c and this c will cancel. So, here you will get 0.8 multiplied by 0.75 e naught prime, just sort of calculate the simple number. You will get 13 by 12 e naught prime. So, what we will find out that an observer in s double prime frame of reference will notice an electric field of 13.12 e naught prime. So, let me write it here. In s double prime frame of reference e y double prime turns out to be equal to 13 by 12 e naught. Let us look at e z double prime. e z double prime is here. Gamma e z plus v v y. e z is 0 in s frame. v y is 0 in s frame. So, e z double prime is equal to 0. Therefore, the only field, only, only component of electric field that you see is only along the y direction, which happens to be 13 by 12 e naught prime, which is of course, larger than what you have seen in s prime frame of reference. But as I have said, this is not the complete story. I should also calculate the magnetic fields. Now, let us calculate the components of the magnetic field s double prime can now be found out using the field values in s because I am using a transformation from s frame. Along the x direction no problem, v x double prime is equal to v x is equal to 0. Let us look at the y component. I am using a direct transformation. Remember, this is equal to gamma v y plus v by c square e z. v y is 0, e z is equal to 0. This is 0. So, x and y component of the magnetic fields are 0, as they were also 0 in s frame. Now, let us calculate v z double prime, the z component of the magnetic field in s double prime frame. v z double prime is equal to gamma v z minus v by c square e y using the direct transformation. Gamma is equal to 5 by 3. I have put 5 by 3. The z component of the magnetic field we had calculated is 0.75 times e naught prime divided by c. Electric field we have calculated 1.25 multiplied by e naught prime by c. This is a relative velocity 0.8 c. This c cancels with one of the squares. So, this just remains e naught prime by c. Just calculate this number. This turns out to be equal to minus 5 by 12 e naught prime by c. So, we see that now in s double prime frame, the magnetic field appears to be in minus z direction because there is a negative sign here. So, an observer in s double prime will feel that in his field, there is a field, electric field in the y direction and a magnetic field in minus z direction. Let us calculate the forces and see whether they obey the transformation law. So, you have said v the c that s double prime has the magnetic field. v the c that in s double prime, the magnetic field would be felt in negative z direction. In order to calculate the force on charge q in s double prime frame, we have to find the speed of s double prime because what has been given are the speeds of frame s prime and s double prime in s. What has also been given is that the charge is at rest in s prime frame of reference. I do not know the speed of the charge. See, from s to s prime, I can find out because if the speed of the charge is point of the s prime frame is 0.6 c and the charge is at rest in s prime, therefore, obviously u is equal to 0.6 c. But here in this particular case, I have to find out by using a velocity transformation. See, I know what is the speed of the frame s double prime in s. I also know what is the speed of the charge in s. I have to find out what is the speed of s in s double prime. I have to use a velocity transformation. Information about the speed of the charge and the speed of the frame has been given to us in s s frame. If I want to find that information in s double prime, s double prime frame, because I have to calculate now the force in s double prime frame of reference and because there is a magnetic field in that frame, therefore, force will depend on what is the speed of the charge carrier in that frame. I must find out the speed of the charge in s double prime frame of reference and for that I must use a velocity transformation, which I am using in the next transparency. This is the standard velocity transformation equation, which we have used in many of the examples earlier. u x minus v divided by 1 minus v u x by c square. See, remember u x is the x component of the particle of the speed of the velocity of the particle. The particle is the charge, which moves with a speed of 0.6 c in s frame of reference. So, u x is equal to 0.6 c. v is the relative velocity between the frame. I want to transform from s to s double prime and that speed is 0.8 c. So, v because 0.8 c divided by 1 minus v, which is 0.6 c, I am sorry 0.8 c, u x which is 0.6 c, c square will cancel with the c square, you get 0.6 multiplied by 0.8, which is 0.48. So, this because 1 minus 0.48, which will give you minus 0.2 divided by 0.52. So, according to an observer in s double prime frame, the charge is moving in minus x direction, which is sort of sensible, because in s frame, the frame s double primes moves with a larger velocity than s prime. Therefore, whatever is at rest in s prime would appear to be moving backwards. Therefore, it is expected that u x double prime will turn out to be negative, which it need turns out to be. Now, as far as the y and z components are concerned, they will give me 0. So, I need not bother about it, because u y is 0, u z is 0. As far as the charge is concerned, it moves only along the plus x direction in s frame. These components are 0, only the x component is non-zero. I do exactly the same equation. I use the trans the Lorentz force equation in s double prime frame of reference. I write the force, force equal to q times e, which I have just now calculated 13 by 12 E naught prime. V, which I have just now calculated is minus, let us remember, you know that this was minus a speed, this was minus 0.2 divided by 0.52. But the magnetic field was also minus 5 by 12 E naught prime by c. So, these two minus makes it plus. So, this becomes 0.2 divided by 0.52 c in i direction cross 5 by 12 E naught prime by c in k direction. I take the cross product i cross k, this will give me minus a. So, here there will be a minus sign. Then you just simplify, multiply these particular things and you get f double prime is equal to 12 by 13 q E naught prime j. So, in s double prime frame of reference, the charge, the force on the charge will be given by 12 by 13 q E naught prime j, which remember is smaller than q E naught prime. Remember, we have said that this force is seen to be largest in a frame of reference in which the charge, the particle happens to be addressed. So, as you can see that these two forces are smaller than the force that this particular particle would have experienced in s prime frame of reference. Now, let us confirm that this follows from the force transformation law. So, let us look at the transformation of the x component of the force. Here, when I am transforming from s prime to s double prime frame of reference, particle is neither addressed at s prime frame of reference nor at s double prime frame of reference. So, I cannot use those short equations. I have to use the complete equations. And complete equations as far as the x component is concerned is here. There is little more involved, but nevertheless we can do because I know all the speeds. So, let us calculate as far as f x is concerned, this is 0. Let us look at f dot u, f dot u will be f x u x plus f y u y plus f z u z. So, I am trying to calculate it in the next transformation. As we have seen, f x is equal to 0. The charge also moves in s frame only along the x direction. Therefore, u y is equal to 0. The force is only along the y direction. So, f z is also equal to 0. Velocity is only along the x direction. So, u z is also equal to 0. So, remember f x u x will make this particular term 0, f y u y because of u y being 0, f y u y also becoming equal to 0, f z being may be equal to 0. Of course, u z is also equal to 0. The dot product f dot u will be 0. So, this will give me the x component of the force to be equal to 0. This anyway I had expected. It is the y component which is more interesting which let us write it here. Complete equation f y divided by gamma 1 minus v u x by c square as far as f y is concerned. This is 0.8 q e naught prime. This is what I have written as far as the frame s is concerned. u x the speed of the charge is along the x direction which is 0.8 c. Gamma we have already said is 5 by 3. So, in this particular equation gamma is 5 by 3. So, I will write this as 3 by 5 because this is 1 upon gamma. So, this is 3 by 5. For f y, I have written 0.8 q e naught prime. So, this is 0.8 q naught prime divided by 1 minus v into u x which is 0.6 into 0.8 which you must 0.48. Just substitute you will find that the y component of the force indeed turns out to be 12 by 13 q e naught prime as we have seen it earlier in this piece of paper. I have written that f double prime is 12 by 13 which I also get from the force transformation equation which I had expected. Now as far as f double prime is concerned, this anyway turns out to be equal to 0. We thus see that the force transformations are obeyed as was expected. So, I am just summarizing all the fields and the forces that I have got in all the three frames. This is an s frame from which we had started. We had written the speed of the particle was 0. B, the magnetic field was 0. The electric field had only y component which is e naught prime. The force on the particle was q e naught prime. Other electric field and the force components were 0. If you go to s frame of reference, things change. u x, the particle is no longer at rest. It is speed is 0.6 c. The particle also experiences a z component of the magnetic field which is in the plus z direction, the magnitude of which is 0.75 e naught prime by c. It also experiences an electric field in the y direction which is now somewhat larger than what was seen by s frame, s prime frame. This is equal to 1.25 e naught prime and the force on this particular charge will turn out to be 0.8 q e naught prime which is the complete Lorentz force, the force which is because of electric field as well as because of the magnetic field. Other velocity fields and the force components will all be 0. Now, let us write in s double prime frame of reference. In s double prime frame of reference, I have to calculate the speed of the charge by using the velocity transformation and that I found out to be equal to u x double prime was equal to minus 0.2 divided by 0.52 c. In this particular frame of reference, there was a z component of the magnetic field but it was pointing out in the negative z direction which was minus 5 by 12 e naught prime by c. And there will be an electric field as usual in the y direction and the magnitude of that will be 13 by 12 e naught prime. So, as we have seen that the electric field also has changed, magnetic field also has changed and the force on the particle is 12 by 13 q e naught prime. The forces is also changed but as we have seen that these forces obey the force transformation law as it was expected. So, I said other velocity field and force components are 0 in this particular frame of reference. So, this is just a summary of what will be the fields which will be seen as s double prime frame of reference and also the forces. As we can see that a field which happened to be purely electrical turns out to be electrical and magnetic, a combination of electric and magnetic field both in s frame and s double prime frame. As you can see the picture is very different. So, there is nothing like a pure electric field, nothing like a pure magnetic field. It depends on the frame from which you want to look into, it could be a combination of them. Before we close, let me do one more thing. Let me try to make a transformation directly from s prime frame to s double prime frame of reference. That is what we have said that where I have to calculate the field in s double prime frame of reference, either I could have started from s or I could have started from s prime. What I did, I started from s frame to s double prime frame of reference. The advantage here was that I knew the relative velocity between the frames but I did not know the charge velocity. In s prime to s double prime, if I calculate, I will not know the relative velocity of the frames because that has not been given to me. I have to use my velocity transformation and find it out. But the charge I know is at rest in s prime frame of reference and the advantage is that because the charge is at rest, so I can use that simplified the force equation. Let me just do this particular thing quickly. So, before we close this example, let us attempt a transformation from s prime to s double prime frame without going through s. So, if I go from s prime frame to s double prime frame of reference, I must know and if I want to use the direct transformation, I must know the speed of s double prime frame of reference in s prime frame of reference. As I said for this, I have to evaluate gamma between s prime and s double prime frame. For this, we have to find out the speed of s double prime in s because I am transforming from s prime to s double prime frame of reference. So, I must know what is the speed of s double prime in s prime frame of reference and information being given in s prime and transforming to s double prime, I would use a direct transformation. However, I need not calculate this because I have already calculated this in a way. I was told that the charge at rest in s prime frame of reference, we have already calculated the speed of the charge in s double prime frame of reference, which was minus whatever it is. So, remember what is the speed of the charge in s double prime frame will also be the speed of s prime frame in s double prime frame of reference because the charge is at rest in s prime frame. So, therefore, whatever I had calculated 0.2 or 0.52 will be the speed of s prime frame in s double prime frame of reference. But if I want to use direct transformation, I must know what is the speed of s double prime in s prime frame of reference, which will just be opposite to the sign. Hence, the speed of the s double prime in s prime frame will be given by v is equal to plus 0.2 divided by 0.52 c because when I had used minus, that was the speed of s prime in s double prime. I want to find out what is the speed of s double prime in s prime, which is just opposite to the sign. If I am going this way, the speed is like this. If this is at rest, the speed is the other way. It is as simple as that. I calculated gamma using this. This turns out to be 13 by 12. Now, I can make electric field transformation as far as x component is concerned, it is straight forward 0. I am using direct transformation. I use gamma is equal to 13 by 12, E y prime minus V B z prime. This will just be given by 13 by 12 E naught prime as I expected. E z prime will turn out to be equal to 0 as I expected. So, I could have directly gone from s prime to s double prime and calculated electric fields without any mistake. Magnetic field, s component is 0, y component is also 0. Let us go to the z component. Z component is V z prime, which happens to be 0 minus V by c square E y prime. V is given by 0.2 divided by 0.52 c. This c will cancel here. The field as far as s prime frame is concerned is just E naught prime. I put it E naught prime. Just work it out. I get exactly minus 5 by 12 E naught prime by c as I had expected. So, I could have gone from s prime to s double prime also or s to s double prime makes no difference. We thus get the identical result as expected. So, in conclusion, we had used another method to derive perpendicular magnetic field transformation and also worked out an example of electric and magnetic field transformation. Thank you.