 So, we have now come to the end of the course. This is going to be the my last lecture trying to wind up whatever we had started about the electric field and magnetic field. In our last lecture, we had taken one example and discussed the electric field and magnetic field transformation. We had said that what is what appears to be a purely a magnetic field or an electric field in a frame may appear to be combination of electric field and magnetic field in a different frame. So, whether you term a field as electric or magnetic or partly electric or partly magnetic could depend on the frame of reference from which you are observing. Today, what we will try to do is try to analyze this situation little bit more and try to see how we can think about the presence of these fields or change these fields by taking one simple example of a current carrying wire or current carrying conductor. Then eventually we will show I mean I will not really be able to show because you know this is actually the beyond the scope of this particular lecture series that Maxwell's equations which are supposed to be the basic equations in electrodynamics they remain unaltered even after special theory of relativity. So, there those equations do not change once the special theory of relativity is also introduced. So, they are invariant under special theory of relativity. So, as I said this particular aspect we will not be able to prove as it is beyond the scope of this thing in this particular course, but we will just take one equation and try to sort of convince I would say try to convince that for one equation we will sort of say that you know you can sort of see that Maxwell's equation is expected to be obeyed. So, this is what we are going to do in our last lecture today. So, this is what we have said we recapitulate we discussed an example using electric and magnetic field transformation in our last lecture. To get a physical feeling how the fields could change in different frames we discuss a new concept today which we call as a current density 4 vector. As I said we want to discuss little bit more physically by taking current carrying conductor, but before we come to that situation we will define what we call as a current density 4 vector. Before we do that let us first take traditionally what we mean by a current density. In physics generally current density is considered little more fundamental than the current and current density is always a vector quantity. Because this current concept of current density emerges more from the fundamental aspects of how the motion of charge carriers takes place in a particular wire or by a particular conductor. So, let us assume that we have n charge carriers per unit volume in a given conductor or given wire which is carrying current. Now of course we know that it is actually the electrons which carry the current which are negatively charged particle. But somehow traditionally the direction of the current has been always defined to be the opposite direction of the flow of the electron. So, if the electrons sort of flow in the negative x direction we define the standard traditional current in the positive x direction. So, as we said let us assume that there are n charge carriers per unit volume and each one of them is carrying a charge of plus e. So, I am just assuming that the charge carriers are positive though as I have said that actually the charge carriers in a real conductor is supposed to be electrons which are negative charge carriers. But just for defining the concept of current density we I take this as a positive charge carriers. Now we evolve the concept of what we call as a drift velocity. Actually once we apply electric field these charge carriers start accelerating under the influence of the force which is created by this particular electric field. But because of this scattering that these electrons or these charge carriers whatever they are face within a metal or within a conductor they are not able to move too much ahead they go little bit ahead and then they get scattered then they again accelerate and then they get scattered. And if you remember what we can define that under the influence of electric field these electrons will eventually develop what we call as a drift velocity. It is something like the motion of air breeze. The molecules in air may be moving with very large speeds but we do not feel it. Only when the overall air drifts which may be drifting with a very small speed in comparison to the actual speed of the molecules that we feel that there is a breeze. So, that is what is the drift of this whole air. So, similarly is the concept of electrons. The electrons could be moving with very large speed inside a conductor. But if there is one electron going this way another electron is going this way. So, eventually you do not see any drift. Only when there is a net flow of electron in a particular direction then only we see the current and that happens when we apply an electric field and then we define what we call as a drift velocity. Now current density vector J is defined in terms of this drift velocity as follows J is equal to n e times drift velocity whereas as I have said n is the total number of charge carriers per unit volume e is the charge of each of the carrier which I am assuming at the moment to be positive. Actually it makes no difference if you take per electron we have to take care of science properly. Otherwise this equation is the equation which defines the current density inside a metal or inside a conductor. Now if this situation is viewed from a different frame then obviously you expect that the velocity of the electrons would not be same because we have discussed number of times that the velocity is actually frame dependent quantity. So, if I go to a different frame the drift velocity that you are seeing in a given frame may turn out to be different. It is very well expected. Also I would like to emphasize that even the current density or charge density not sorry current density the charge density would be different and that is mainly because of the length contraction because once we have come to the case of relativity theory you will find that there is a contraction of length and that itself would change the charge density. And if charge density change and the velocity drift velocity change so when I look into a different frame of reference the current density is also expected to change. Therefore I must look on in relativity how this current density will change if I change my frame of reference. That is what is the question that we are going to answer now. This is what I said let us examine this concept from the point of view of relativity. For that particular thing let us assume comparatively simpler picture. Let us not bother about the drift velocities and other things. Let us just take comparatively very very simple picture and try to see how these current densities will transform if I go from one particular frame of reference to another frame of reference. So, let us assume that we have a volume which contains only positive charges each one of which has a value of q which moves with a velocity u. So, let us assume that all of them move in the same velocity because there is no question of drift here. I do not want to confuse. We will talk about drift little later. So, let us assume that they all of them are moving with the same velocity u as seen in a given frame S. Obviously, when these charges move in a particular direction this will constitute current or eventually current density. Let N be the number of charge carriers per unit volume as seen in this frame. We have just now said that this N is also likely to be frame dependent. I will discuss this point little more in detail little later. So, we have a frame as we have been always telling we have to be consistent in our frame. So, there is an observer sitting in S frame which observes the velocity of the charge carriers to be u and once he calculates the total charge number density or charge density. If you let us talk of number density then that is the number of charge carriers per unit volume. He observes that number to be equal to N. Now, let us define a 4 scalar. We have introduced the concept of 4 scalar earlier. This is the quantity which does not change when we change our frame of reference. See like we have defined when we from velocity to momentum. At that time we multiplied by a 4 scalar which we called as the rest mass of the particle. Similarly, here we define a 4 scalar N naught which does not change in your frame of reference. This of course is going to be different from N and this is the number density that is the number of charge carriers per unit volume in a frame of reference in which these charges are at rest. See like rest mass we define that this is the mass what we have called in a frame of reference when the particular mass is at rest. Take analogy from that particular concept. Similarly, we define a number density, total number of charge carriers per unit volume as evaluated in a frame of reference in which these charges are at rest. This N that a person is going to find out in S is going to be different from N naught as we will see just now. Now, the charge density in the frame S is higher from the proper one. This number N I can call as a proper number density and not sorry this N naught we can call as a proper number density because this is evaluated in a frame of reference in which these charges are at rest. Now, what I insist that the N that I am evaluating in S, S is a frame in which the charges are actually moving. In this frame the number density would be different from its proper density N naught mainly because of the length contraction and this length contraction always occurs along the relative velocity direction. So, let us just look into this particular picture. Let us suppose we have a wire or a conductor like this and let us assume that this is x direction. Now, there are charge carriers here. Let us say negative charge carriers whatever are here we have assumed positive. So, let us assume positive. Let us assume that they are moving in a particular direction x. Now, if I view this particular thing from a different frame of reference then in that particular case and let us assume that this particular frame of reference let us say S prime or whatever is that particular frame of reference views a person sitting in that frame views these charge carriers then that particular person would find that this length is contracted. I am assuming that this particular frame is actually moving in the same direction as x. So, we know the standard length contraction formula. So, according to this person the length will turn out to be smaller, other dimensions will not change and this length will become smaller by a factor of gamma u. New length or the length as seen in different frame of reference will be gamma u times the original or the proper length. Therefore, the volume of this particular material will also go down by gamma u because no other dimensions change. So, if you have let us say you know the dimensions are a, b, c then only one of them becomes gamma a and b and c remain same. So, when I take the volume, volume goes down by a factor of gamma u and therefore charge density the total number of charge carriers per unit volume go up by a factor of gamma u. So, that is what I say that if I have come to this particular frame of reference s in which I find that these charges are moving then this number density that will be evaluated in frame will be different from its proper number density because of the length contraction and therefore this particular factor will get modified or get multiplied by a factor of gamma u. Here is the current density if I have to write in s frame. I have to write the total number of charge carriers per unit volume which I know I have earlier written as n will actually be gamma u times n naught where n naught is actually a full scalar. This factor is n and this n is gamma u times n naught using the definition of charge density that we have used earlier. This equation we can write that current density is total number of charge carriers per unit volume which is gamma u times n naught multiplied by h q multiplied by the velocity I am assuming all of them are moving in the same velocity. So, this will be the current density as seen in this particular frame of reference. So, the only way of writing only reason why I am writing this equation in this particular form because I am using a full scalar here. If I would have written just n this n would have changed if I want to different frame of reference while this n naught is going to be same in all the frames. Now, instead of number density sometimes we define charge density. This charge density is very simple just multiply whatever is n naught multiplied by the charge which is q. So, this is a very simple definition most of the time we talk in terms of charge density rather number density. So, we define the charge density rho and like we have defined a proper number density we can define a proper charge density rho naught as follows. So, rho naught whatever was the number density multiplied by charge is very simple just charge density the total number multiplied by charge. Similarly, the charge density in a particular frame is the total number density as seen in that frame multiplied by q. So, this rho is equal to n q and rho naught which is the proper charge density is n naught q and because as we have discussed n is equal to gamma u n naught. So, this particular factor rho naught this factor this equation rho can be written as gamma u rho naught because this is n naught q this n I can write as gamma u n naught n naught if I absorb here this becomes rho naught. So, this becomes gamma u rho naught. So, rho charge density in a particular frame of reference can be written as gamma u rho naught and only thing which I insist that this rho naught is also a 4 scalar because n naught is a 4 scalar and q the charge in relativity we do not expect this to change once we change the frame of reference. At this point we have not specifically mentioned earlier, but let us emphasize it now the charges do not change once we change the frame of reference. Therefore, this quantity rho naught is also a 4 scalar. So, we can write the current density now in terms of this equation gamma j is gamma u rho naught u this is what we have written here earlier except this gamma u n naught this n naught q's I am sorry have been written in terms of rho naught that is all I have done in this equation. This is j is equal to gamma u rho naught u now I am in a position to define current density 4 vector. If you remember in one of our earlier lectures we have talked about the velocity 4 vector we have said that u x, u y, u z they are not the components of the 4 vector or the velocity 4 vector. So, remember p x, p y, p z are the first 3 components of the momentum 4 vector, but u x, u y, u z are not the components or not the first 3 components of velocity 4 vector. In fact, what are actually the first 3 components are gamma u u x, gamma u u y, gamma u u z this aspect we had discussed much more in detail in when we are discussing the velocity 4 vector. So, this velocity 4 vector the components are given as gamma u times u x this is the first component, second component is u y, third component is u z this is the I c all have to be multiplied by gamma u. So, strictly that means that the first component of velocity 4 vector will be gamma u u x, second will be gamma u u y, third will be gamma u u z and the fourth component will be I gamma u c. This is what we have defined earlier. Now, I can use this particular thing to define charge density 4 vector all I have to do is to multiply by a 4 scalar which I can do by multiplying by rho naught because we have earlier said that rho naught is actually 4 scalar. So, we define the components of current density 4 vector by multiplying this u 4 vector by a 4 scalar which is rho naught. So, this has been multiplied by rho naught all other components are exactly same only thing this has been multiplied by rho naught. Now, we realize if you look at back at the equation of current density this I can write component wise if I write component wise it will become j x whatever x component will become j x gamma u rho naught u x the y component of this equation becomes j y is equal to gamma u rho naught u y things like that. So, once I write here rho naught gamma u u x this becomes j x once I say rho naught gamma u u y this becomes j y rho naught gamma u u z this becomes j z and I am retaining this particular thing writing this rho naught gamma u as rho which is the charge density this becomes I times rho multiplied by c. So, what we see that the first three components of current density 4 vector are actually the real current density as seen in that particular frame it is like momentum, momentum 4 vector the first three components are the momentum as observed in a particular frame of reference. Similarly, current density components are the first three components of the current density 4 vector and the fourth component like in the case of momentum 4 vector dependent on the energy of the system here the fourth component of the current density 4 vector depends on the charge density. So, the fourth component is rho see like in that particular case of momentum it was I e upon c the fourth component dependent on the energy here the fourth component depends on the charge density. Now, once we have found this particular 4 vector we know how to transform it if I go from this particular frame of s to any other frame of s and if I know the relative velocity between s and s to be v I can always transform this particular 4 vector components into any other frame by using the standard matrix equation which we have been doing earlier a number of times. So, the current density will obey the following transformation rule if we go to another frame s and this is the current density transformation. So, we have in a s frame these are the components the current density is J x J y J z these are the x y and z components of the current density the fourth component is I rho upon c where rho is the charge density as seen in this particular frame of reference I want to find out the current density and the charge density in a different frame s frame of reference and let those current densities begin be given by or the components of the current densities in s be given by J x j y j z and let the charge density be given by rho then these J x prime J y prime J z prime rho prime would depend on J x J y J z and rho by this particular matrix equation transformation equation where of course this gamma beta all these depends on the v the relative velocity between the frame between s and s prime the standard way we have been dealing in relativity so far. So, now I can expand this particular thing we have done it a number of times so we will not spend too much of time in expanding this particular thing I will just write the expansion of this particular equation immediately in the next transparency this is what it becomes J x prime becomes equal to gamma times J x minus v rho J y prime remains equal to J y J z prime remains equal to J z and the charge density now will be different in a different frame s prime as given by this particular expression rho prime will be gamma times rho minus v J x upon c square this is somewhat like your x y z if you want to remember somewhat like x y z t transformation x prime turned out to be gamma x minus v t so it is a similar type of equations if you go back in that particular case J x prime is gamma times J x minus v rho and rho prime is rho gamma minus v upon c square J x I can always write the inverse transformation equation for that particular thing as we have said change prime to unprimed quantities and vice versa make v minus v so this what happens as the inverse transformation it means if I know the current densities and the charge densities in s prime frame of reference I can find out the current density and the charge density in s frame of reference by using this these inverse transformation equations so this is what has happened to my transformation of current density now let us come back to the situation of current carrying conductor my idea of using this particular example of giving current carrying conductor is only to look little more little more deeply into the origin of these electric fields and magnetic fields as we change from one frame to another frame of reference and this particular example generally reasonably illustrative because many times we have a question that suppose there is a particular frame of reference in which we see only magnetic field and you say that you go to a different frame of reference now in addition to this magnetic field you also have an electric field so from where this electric field has come because once we say that equations the Maxwell's equations which are the basic equations do not change in relativity so what is the origin of these electric field so let us what we are trying to do by giving this particular example to go little deeper into this particular aspect try to somewhat understand this thing so now let us come to a realistic conductor and let us assume that it has it is carrying a current in plus x direction as seen in frame s so there is a frame s in which you see a current flowing in a particular wire or conductor along the x direction let us assume that charge carriers are actually the electrons it means actually the electrons must be drifting in minus x direction and that is the reason that the current is actually in the plus x direction as we have said the direction of the motion or direction of the drift of the electron is opposite to the direction of the conventional current. Now I want to I mean when we talk about the current we do not talk about this so clearly but let us be very very specific I am looking for this particular thing from a frame s and in this frame s I assume that the positive charge carriers remember the conductor overall is electrically neutral it is a slightly different situation from what we have discussed just now I am talking of now a realistic conductor if there was no electric field this particular conductor was electrically neutral and their negative charges and their positive charges both are there now these electrons tend to be somewhat free or approximately free and when we apply these electric fields these electrons are the one which actually give rise to the current but those positive charges which are sort of which are generally believed to be immobile which are ions which have been left behind they remain at rest that is what is the normal picture of current that is always told to us in high speed so I am looking at the same frame in which this picture is correct it means I am assuming that the positive charges are at rest and it is the negative charges which are drifting in negative x direction to give rise to a current in plus x direction that is what I have said here in this frame the positive charges are at rest and electrons drift in negative x direction giving a current in positive x direction alright now let me assume we all of all of us know that if there is a current carrying conductor let us assume that is a very long conductor this conductor will generate magnetic field around it so because of this particular current there is a magnetic field which is created around it we know how to find out the direction of the magnetic field if the current is going like this now you have a you can find out there will be all the magnetic field lines of forces will be circular with you know the wire passing through the center of the circles and these are well known actually let us not discuss them in detail what I am also insisting that is all generating only the magnetic field without any electric field so there is no electric field okay and the reason there is no electric field because there are positive charge carriers as well as negative charge carriers and in a given volume whatever small element that we take we assume that they are same number of positive charge carriers and same number of negative charge carriers so the net let me just write it here if we take a small element in a wire in this element negative charge carriers and positive charge carriers both are same therefore if you take a small section of the wire you do not find a net charge density here and therefore there is no electric field which is generated outside there is only a flow of electron at all the times this charge neutrality is maintained in a small section of this particular wire therefore there is no electric field but there is a magnetic field. So this is picture with which we are quite familiar in our high school then we have been discussing various laws related to magnetism at that time we have been talking about this type of behavior. So this is what I said let the negative and positive charge densities in this frame as b rho e and rho p this e symbol I have said for negative charge carriers and this p symbol I have said for positive charge carriers. So let us assume that the charge density for negative charge carriers is rho e and for positive charge carriers is rho p of course because this is charge density so this is negative because the charge is negative though number density is not negative number density has to be always positive while positive charge density is positive because the charge is positive and because we are saying that there is no electric field in this particular frame of reference I expect that if I take any small unit volume by small unit necessarily small infinitesimally small volume you are always find that this rho p plus rho e is equal to 0. So there is no charge density in a smallest volume a small volume which you take along the wire I am not writing in the differential form etc because that is not the idea we want to make things simple. So just want to say that I expect that this rho p plus rho e must be 0 in this particular frame because I do not see an electric field I see only the magnetic field there is a current there is a current density but there is no electric field this current and current density causes the magnetic field. Now we realize that because in this particular frame it is only the negative charges which are moving so the current density the net current density is being caused only by the motion of negative charge carriers positive charge carriers would not contribute to any current density which is obvious because they are not moving they are not moving in this particular frame of the first. So this is what I have said in this frame S the current density is solely due to the electrons therefore net current density I can write as equal to J x e I am writing only x component because current density current I have been told that is only in the x direction current density is also in the x direction J x e I am using to mention the x component of the current density for negative charge carriers. And as we have said and J x is the overall net the total current density as we have said because positive charges are immobile so this J x is solely because of J x e and that will be given by rho e multiplied by the drift velocity of the electrons or negative charge carriers negative charge carriers we mean electron. This is what it will be given as rho e times u d e where rho is the charge density u d e is the drift velocity we have already defined this particular thing is exactly the same thing. Now let us go to a different frame of reference S in which the drift velocity if electron is found to be 0. So I have already found out what is the drift velocity if I know the current density I know what is the drift velocity if I know the drift velocity I can always go to a frame of reference an inertial frame of reference which has the same relative velocity with respect to S as the drift velocity of the electrons. What it means is that if you have a wire in which these electrons are moving in this particular direction this is plus x direction let us consider one electron which has exactly the same speed let us not bother about that you know let us just consider a drift velocity and assume that overall electrons are drifting in this particular way. I go to a particular frame of reference which has exactly the same velocity of course this frame of reference also has to move in this particular direction. So that they find that overall there is no drift of electrons in this particular frame of reference. So in this particular S frame of reference I am defining this particular frame of reference S as one frame of reference which of course is the inertial frame of reference because the drift velocity is supposed to be constant for a given electric field so long current density is constant the drift velocity is also constant. So I am looking at this particular aspect from a frame of reference S in which electrons are not found to be drifting so that is what I said let us observe this wire from a frame S in which the drift velocity of the electron is found to be 0. Let us see what an observer in S frame would notice would it notice that the current has become 0 and therefore there should not be any magnetic field that is not a current picture. Actually in this frame the positive charges will be observed to be moving in plus x direction see remember it was in S frame that your electrons were drifting and the positive charges were stationary. But once I have changed my frame of reference to make the electron stationary in that particular frame of reference in that particular frame of reference the positive charges will move and because as we have said here this velocity is in the negative direction. So positive charges choose a mobile earlier which would appear to a person sitting here to be moving in positive x direction. Therefore an observer in S frame would feel that all these positive charge carriers are moving in plus x direction. So there is a current in that particular frame of reference but that current is being caused by the positive charge carriers and this positive charge carriers will actually produce a magnetic field. Therefore a magnetic field is expected to be present also in S frame of reference. What about electric field? If you look at these equations these are the transformation equations which I have written from S to S frame of reference. Of course in this case v happens to be negative let us forget about it I am just writing in the number form. So if you look at this particular equation of course ex will turn out to be equal to ex but ey will depend on vbz, ez will depend on vby depending upon which point you are looking or the field at which point of course the current is in the x direction you will always find you may always find at least a component of by or vz or both. Even though ey and ez are 0 because we have said in S frame there is no electric field but there is a magnetic field and this magnetic field has to be in some direction of course it cannot be in x direction because the current flows in the x direction and the magnetic field is being caused only by that current flow and I am assuming this to be infinite wire so sort of infinite wire. So therefore at least one of these will be non-zero it means in this particular frame in addition to the magnetic field the observer would also find electric field this is going to happen if my transformation equations are correct. So these electric fields are present in this particular frame of reference from where are they arising earlier we have said that their charge densities were just neutralizing at each small volume of the wire and therefore there is no net charge in any small section of the wire and therefore there was no electric field earlier from where the electric field is coming. So now we will realize that is what we will do just now to show that that was too earlier in S frame in S frame of reference if you take a small section of the wire now you will find it to be charged and if you find it to be charged it is this charge which will generate this electric field. So electric field will be present in this particular frame of reference because of that particular section I mean sections of wire being charged how do I do that for that I must do a charge density transformation that is what I am going to do next. For that let us evaluate the charge densities in S frame of reference I have just now evaluated I just now found out the transformation equation relating to the charge densities let us do the same thing here. Find out what will be rho prime if I know rho in S what will be rho prime of course relative velocity of between the frames I have taken specifically equal to UDE of course I have reserved symbol U instead of V though when we originally defined charge densities we had used symbol V but I am using U because V we have as I said always reserved for the relative velocity between the frames of course in this specific example I am taking V is equal to UDE because only in that particular case the electrons or the negative charge carriers will be at rest this is my transformation equation for the negative charge carriers rho E prime is equal to gamma rho E minus UDE because V is equal to UDE J XC whatever was the current density divided by C square this is gamma rho J XC I can write as rho E into UDE. So, this rho E I have written here this becomes UDE square divided by C square this gamma this rho E I can take it out this will be the gamma rho E multiplied by 1 minus UDE divided by C square but this is also squared UDE is the drift velocity of electrons as was seen in S frame. So, this will be the charge density of electron which will be seen in S prime frame of reference what will happen to the charge density of positive charge carriers I will write exactly the same equation gamma rho P minus V J XP divided by C square where J XP is the current density of positive charge carriers in S frame but I know that this J XP 0 because the charges were not moving in S frame of reference therefore current density in that particular frame has to be 0. Therefore, this quantity will be 0 and this I can write as gamma rho P as we can see that the charge density is as seen in S frame of reference are not same because of this transformation. Therefore, this is not equal to this they were earlier same their magnitudes were earlier same good signs were different their magnitudes were same in S frame but in S frame of reference even their magnitudes have become different. So, let us evaluate the net charge density in S frame of reference which is just the sum of these two which earlier was 0. So, I have just used this particular equation which is from this particular transparency this plus this this plus this this what I have written here this gamma rho E would cancel out here remember this rho E was equal to minus rho P in S frame of reference. So, this rho E has been changed as minus rho P here. So, once I do that this gamma rho P and this will cancel out and this will become gamma rho P UDE square upon C square. So, we find that there is a net charge density in this particular wire. So, in S frame of reference it will find that the wire is charged if you take one particular section of the wire will be charged and if it is charged of course, we will generate electric field. Now, what happens to charge conservation some of you may ask that does it mean to say that this wire overall has become electrically non-neutral originally we have said that the conductor was electrically neutral. Now, we are saying that there is a net charge density does it mean to say that it is no longer electrically neutral. What happened from where we got the additional charges is the charge conservation obeyed or not the question is that see normally you will not have strictly speaking infinite wire you will always have a loop. So, once you are supplying the current there will be current flowing and then eventually the current has to flow back into a particular direction. So, whatever it is a situation you will always have situation somewhere where you have wire and this wire eventually has to close here there has to be some current source. Now, as you can see the current direction in this section of the wire is going to become different from whatever is here the if I am looking at the positive x direction the charges are going let us say the current is going in this particular direction this way while here it is going this way. So, in one section of the wire if I am positively finding it positively charged the other section of the wire I will find it negatively charged. So, overall the wire will not be will still remain to be electrically neutral. So, that nothing happens to that particular thing must be realized that thing, but if you take a specific section the way we have been defining in this particular thing that we feel only because of this particular wire that comes because of this positive charges which are being in that particular frame which you know which views this particular charge carrier this particular wire as a charged charge. Now, with this particular thing as I said this probably the last section that we have wanted to cover formally I want to just introduce how the Maxwell's equation we expect them to remain invariant. In fact, if you remember the electric field and magnetic field transformation we had obtained from the force equation the Lorentz force equation and saying that this force must obey a force transformation law. In fact, they can also be derived by maintaining that the Maxwell's equations are unaltered or unaffected when I change my frame of reference. We do not change any magnetic equations unlike we have changed many of the classical mechanics equations the equations pertaining to electromagnetics electromagnetic theory they do not change once we change my frame of reference the equations which are the equations Maxwell's equation for Maxwell's equation they do not change. So, I will not prove in general which is beyond the scope of this course which require if you want a general proof much more many more details. We will just take one simple example and convince you know I am always trying to say that I am trying to convince that Maxwell's equations are expected to be moving nothing like that nothing more than that. So, this is what I have said in relativity Maxwell's equations are unaffected we should not give the general proof we are just trying to convince you. So, these are the four Maxwell's equations which are very very standard equations the first equation is essentially what we call as a Gauss's law which talks of divergence of electric field in terms of the charge density. This is a similar Gauss's law for the magnetostatics which says divergence of B is equal to 0 and this quantity is 0 because it always says that there is no monopole. This is what is the curl of electric field which is called Faraday's laws of electromagnetic induction and this is Ampere's law with the correction of Maxwell pertaining to this displacement current which gives you curl B. So, the two equations are having diverges there are two equations which are having curl and these are the four equations which we call as Maxwell's equation. So, I will take one of these equations this particular equation the third equation and out of that I will take one component and try to convince you that this does not alter once I go from a frame S to S. So, this is what I am trying to do in this particular as the last part of this particular course. So, we said we shall take third equation and try to write this in S frame but before we do that let us expand this curl into individual components. So, remember it has a curl here, there is a curl of electric field and here you have derivative time derivative of magnetic field. So, let us first write this into the component form. Then this components will of course be in the form of x, y and z and on the right hand side you will have time derivative. Now, I will change these x, y, z to x prime, y prime, z prime this t to t prime assuming that x prime, y prime, z prime, t prime are related to x, y, z, t by Lorentz transformation because that is what is the relativistic transformation and try to see that I can write exactly similar equation in S frame of reference where e will be replaced by e prime and b will be replaced by b prime and this curl will be all with respect to x prime, y prime, z prime and this t will be with respect to t prime. Once I am able to write that I have shown that this Maxwell's equation remain invariant. Of course, we can do similar type of things for all other equations. We know how the charge density transforms we have just now discussed. We have also known how current density transforms that also we have discussed. So, we know the transformation of all these equations. So, in principle, you could have taken any equation and try to prove it. But as I said, I am not doing that in general just taking this particular equation trying to convince you. So, let us first expand this curl del cross e. So, left hand side of that equation had just curl of e that is del cross e. Of course, you can use any method of expanding it. You know some people try to write in form of determinant then try to expand it. I have written in a particular vector form which I am finding which I find generally much more easy to remember. But in any way you would prefer to write this particular vector one can expand this curl. So, this del operator as we call can be written as i del del x plus j del del y plus k del del z where i, j, k are the unit vectors. Then cross e I can write in the vector form as i, e, x plus j, e, y plus k, e, z again i, j, k are unit vectors in the x, y and z direction the standard way. So, I have to expand this curl it means I have to take this cross this, this cross this, this cross this, then plus this cross this, plus this cross this, plus this cross this, then this cross this, this cross this, this cross this. If I take first term i del del x when take i cross i I will get 0, i cross j will give me k and i cross k will give me minus j. Similarly, j cross i will get minus k, j cross j will give me 0 and j cross k will give me i. Here also k cross k will give me 0. So, all I wanted to say that normally if you would have expanded this, you would have got 9 terms because one has to operate on 3, second also has to operate on 3, third also has to operate on 3, we will get 9 terms. But out of those 3 terms will be 0, one because of i cross i, another because of j cross j and third will be because of k cross k. So, eventually you will be landing only with the 6 terms. I am not giving the details of these things, I can think you can work out simply and try to convince yourself that whatever I have written is probably correct. So, these are my cross products and all I have done is expanded into this particular form. As I have said that this expansion you could have worked out in any other way which is comfortable to you. So, all I have written is 6 terms and contained, kept those terms along the x direction here, y direction here and the z direction here. So, this del cross E I can write like this that this has to be equated to the right hand side which had del B del t, the x component and y component and z component of this equation in fact can be written as i del B x del t plus j del B y del t plus k del B z del t. Then this x component can be equated to the x component of the curl that we have obtained, y component can be equated to the y component of the curl that I have found out, z component can be equated to the z component. So, I will get 3 equations which I am writing in the next transparency. So, these are collecting them, we have collected all the components and equated to the right hand side. So, I get these 3 equations. Now, I will use only one of this equation, this particular equation, the second equation, try to transform it into S frame of reference by using the standard equation corresponding to the partial derivatives. I will not go into the details because this requires lot of mathematics as I say my idea is only to convince you and not to give you too much of details of the mathematics which I say as the beyond of the scope of this particular course. So, we shall use the standard partial differential formula and use it along with the Lorentz transformation. What is this particular formula? This del del x can be written as del x prime del x del del x prime plus del y prime del x del del y prime del z prime del x del t prime del x del del t prime. This is the standard partial derivative formula when you have there are 4 variables on which it is dependent. I know from Lorentz transformation x prime is equal to gamma x minus vt y prime is equal to y z prime is equal to z t prime is equal to gamma t minus vx upon c square. So, let us examine this particular term. If I take del x prime del x, this is x prime, this is partial derivative. It means all other variables I have to take them as a constant. So, t will turn out to be constant. Once I take the partial derivative of x with respect to x prime with respect to x, so I will just get gamma. This term will give me 0. So, this becomes gamma into del del x prime. Similarly, del y prime del x, if I take partial derivative with respect to x, it means y has to be treated as constant. So, this will give me 0. Del z prime del x will give me 0. Here del t prime del x, t prime does depend on x. Once I take partial derivative, t has to be taken as constant. I take derivatives, I will get minus gamma t v upon c square. So, this is gamma, not t minus gamma v upon c square. So, minus gamma v upon c square into del del del t prime. So, this operator del del x can be replaced by this operator once I change my frame from s to s prime. I can write similar equations for y which will give me very clearly del y is equal to del y prime. This you can try to work it out yourself and exactly similarly I will get del del z is equal to del z del z prime. Again del del t will be different from del t prime. I use exactly the same thing. Again you can work it out yourself. Del del t will turn out to be equal to gamma del del t prime minus v del del x prime. So, now I know how to replace this partial derivative from s frame to s frame of reference. So, this things I will substitute in this particular equation. So, this del del z, del del z I will replace by whatever we have evaluated earlier. Of course, as far as del del z is concerned, it is just z becomes z prime. Here there is a partial derivative with respect to x. So, this once I change to del x prime, I will get another term. Similarly, when this partial derivative with respect to time, I will get another term which I am doing in the next transparency. I am rather going fast, but idea has to say is to convince you. So, this equation as I said remains identical. Del del x, I am replacing by this equation. So, this once I go to prime frame of reference, this becomes del e z del x prime minus v upon c square del e z del t prime. This del t also gets changed and this becomes del v y del t prime minus v del v y del x prime. So, this equation now becomes in this particular form. What I will do? I will collect terms corresponding to partial derivative with respect to x and with respect to partial derivative with respect to t. So, if I do that, this equation becomes in this form. This is interesting because normally if a similar equation has to be obeyed in s prime frame of reference, I expect del e x prime is equal to del z prime minus del. This is an equation which I expect to be true because this is exactly the same equation. I am sorry, there should be a prime here because this is what I expect. If the same equation I am able to write in s prime frame of reference, I will be able to say that things are consistent in s prime frame of reference. This is the equation which I have written here. So, if I write this equal to, if this happens to be equal to e z prime and if this happens to be equal to b y prime, I know that these equations will be identical in s prime frame of reference. This is precisely what I know is true from electric field and magnetic field transformation. Of course, e x must be equal to x prime. So, from this, I can write these equations. So, these equations will be consistent in s prime provided we have these three equations valid. I know from electric field and magnetic field transformation that these equations are correct. So, I expect this equation to be valid also in s prime frame of reference. So, what I have done is shown only for one equation. You can try to show for other equations. You have to do somewhat more mathematics and try to convince yourself that Maxwell's equations are actually consistent in all the both the frames. They remain unaltered. We do not change Maxwell's equations once I go from s frame to s prime frame of reference. So, this is my summary. We discussed current density four vector. We try to analyze the current carrying conductor in two frames, especially discuss how the electric field is generated in that particular frame of reference and given an hint how Maxwell's equation remain unaltered. So, this happens to be the end of the course. Best of luck, all the best.