 In the last couple of lectures, we have been talking about the various quantities associated with electromagnetic field. For instance, we found out that electromagnetic field carries energy, it carries momentum. And today, we will try to talk about another application that is electromagnetic field also carries angular momentum. And in fact, we will see some very interesting consequences that might arise because of conservation of angular momentum. We had introduced what we called as a Maxwell's stress tensor. And in terms of which, we had tried to explain the conservation of linear momentum. And let us, so in today, the first part of the lecture, I will be talking about angular momentum and its conservation with respect to electromagnetic field. And later, we will go over to an application of time dependent phenomena, namely propagation of electromagnetic waves. As always, we have this workhorse before us. The four equations known as the Maxwell's equation, the del dot p is rho by epsilon 0, del dot of b equal to 0, then Faraday's law, del cross equal to minus d b by d t, del cross h, the ampere Maxwell's law given by J free plus d d by d t. In addition, we have constitutive relations of this type, d equal to epsilon 0 e plus p, h equal to b by mu 0 minus m. In addition, we have things like, for instance, the continuity equations or for instance, if you are talking about a metal, may be Ohm's law is valid and things like that. So, we will be talking about this as we go on. This was our expression for the Maxwell's stress tensor, in terms of which, we had defined the momentum density of the electromagnetic waves. Basically, it has two terms. One is for, from electric field, the other is from the magnetic field and the expression is fairly simple. So, let us look at, how does one get, define the angular momentum. The angular momentum, of course, if electromagnetic field has a momentum, then it stands to reason that about an origin, let us take any origin. I should be able to define the angular momentum by the standard relationship, namely l is equal to r cross p. So, since I am talking about the densities, I am using small letters. So, this is angular momentum density, which is equal to r cross the momentum density of the electromagnetic field. So, what we will do is this, that rate of change of angular momentum, this is all in terms of the density. This is equal to r cross, remember that angular momentum's change is related to the torque that acts on it. So, therefore, r cross the force, as usual this is rho e, that is the force exerted by the electric field and of course, j cross b, which is the usual Lorentz force equation, j cross b. So, therefore, if I, I had already shown how this expression is simplified by going over to Maxwell's stress tensor. And so, as a result since I have r cross everywhere, I will be able to borrow this expression literally. So, therefore, what I would get is d by d t of the angular momentum density plus 1 over c square r cross s, if you recall that I had 1 over c square s, s is the point in vector. So, I have 1 over c square r cross s and that was equal to the part, which ultimately translated as the Maxwell's stress tensor. So, I had epsilon 0 r cross e del dot e, I am not going to repeat that algebra, minus e cross del cross e and a corresponding term from the magnetic field, which is r cross b del dot b minus b cross del cross e. Now, we had already simplified this term before. So, therefore, this would lead to an equation of this type, d by d t of l, this I will now put in mech, because this is mechanical angular momentum plus the electromagnetic angular momentum. These are the total change in the angular momentum of the sources and the electromagnetic fields. So, that is equal to r cross, if you recall, this term was divergence of the second rank tensor, namely Maxwell's stress tensor. So, this of course, s is s by c square, this is my electromagnetic angular momentum and this is r cross the divergence of the stress tensor. So, this is what we had seen. So, if you now integrate it over the complete volume, what you would get would be that statement of conservation of angular momentum and now that it is being integrated, the only dependence is through time. So, therefore, this is d by d t l mechanical plus integral over the volume of l electromagnetic, that is the momentum density d cube x and this change results in if you like a flux of torque through the surface, because this is we know that rate of change of angular momentum is related to torque and this is of course, the total angular momentum. So, therefore, this would be a surface integral, the derivation is exactly the same as before and this is r cross Maxwell's stress tensor dotted with d s. The interpretation as I said, the left hand side means what is the total rate of change of the momentum of the combined system, namely the angular momentum of the source and angular momentum of the electromagnetic field and that can change only if there is a flow through the surface or if you like it is a flow of torque. There is an interesting problem which I would be doing and this is known as the Feynman paradox. Feynman had talked about this, we are going to be using a very similar problem. So, the problem consists of the following. I have a line charge, infinite line charge which has a charge density minus lambda. Surrounding this, surrounding this is a cylindrical surface and this cylindrical surface contains a charge density, surface charge density which is lambda which is the same as the magnitude of this one divided by 2 pi a namely the circumference. So, of course, this then has the dimension of charge density. Now, the question is this that in this situation there is electric field only within say if this radius is a then only between 0 and a there is electric field that is because we have adjusted this charge density in such a way that this is equal and opposite to the charge density of this line charge. So, as a result outside if you use your standard Gauss's law the electric field is 0. So, the question is this that when supposing in through this I have a current is flowing and I decrease the current to 0 this. So, if I decrease the current to 0 then you will find this cylinder will start rotating they how does it happen that the cylinder starts rotating. So, this is this has a charge density lambda and there is a current density J. So, let us look at why does it rotate the Feynman paradox was because the since the system was initially at rest it should continue to be at rest because otherwise it will violate angular momentum conservation. The reason is that when you talk about the conservation of angular momentum you have to also take into account the angular momentum of the electromagnetic field and that is what we are going to be talking today. So, let us look at what we said we said that angular momentum L was this is I am talking about electromagnetic field angular momentum only in my initial angular momentum of the mechanical system was 0 because everything was at rest. So, this is r cross the pointing vector by c square and since it is total I get 2 pi r d r and integrated over the I know what is the electric field. So, electric field is minus lambda divided by 2 pi epsilon 0 r and I know and of course, it is along the radial direction and this is valid for r less than a only there is no electric field outside because the charge densities have been adjusted that way. So, let us look at what does it give me for the momentum charge density. So, momentum density is 1 over mu 0 c square what I am doing is to write down an expression for s if you recall is e cross h. So, h is written as b by mu 0. So, therefore, this is e cross b this is and that is equal to 1 over mu 0 c square the minus lambda divided by 2 pi epsilon 0 r r and the external magnetic field which we have put in let us say there is an external magnetic field along the z direction which is b 0 z. Now, what I can do is since I have a cylindrical geometry. So, I have got r phi z. So, therefore, this quantity gives me mu 0 well c square and there recall mu 0 time epsilon 0 is equal to 1 over c square. So, that will cancel this c square there I will have to with a b 0 lambda divided by 2 pi r and it is azimuthal because there is a r cross z with the minus sign there. So, if this is my momentum density the angular momentum l becomes let me pull out the constants first b 0 lambda divided by 2 pi integral from 0 to a r cross phi divided by r then 2 pi r d r. So, r cross phi of course is in the z direction. So, this is I have got b 0 lambda divided by 2 pi 2 pi also actually goes because of this there is another factor there and this has a magnitude r. So, r and r goes I am left with r square by 2. So, when integrated gives me a square by 2 and the direction z. So, this is the situation that I have got an angular momentum the field which I have calculated is given by this expression. Now, what happens when we switch off the magnetic field? Now, when I switch off the magnetic field in the presence of the magnetic field there was a flux through the this surface here the any of the cross sectional surface there was a flux and when this magnetic field is let us say slowly reduced to 0 there is a change in the flux. Now, this change in the flux because you are reducing the magnetic field in the z direction from some value v 0 to 0 results in an azimuthal current and this azimuthal current I can calculate. So, this is this is the expression for my angular momentum of the electromagnetic field which actually happens to be the total angular momentum because my system mechanical system was at rest. So, this is this is the initial angular total angular momentum. Now, let us then see what is happening at the end. So, what I have done is I have gradually reduced the magnetic field to 0 which has resulted in an azimuthal current and I can calculate that azimuthal current density J phi which is q by t and this is there is a charge which is being supposing the velocity that has come up the angular velocity that has come up is omega. So, therefore, this is 2 pi a sigma that is the amount of charge that I had into omega by 2 pi and that is lambda omega by 2 pi. So, when it rotates this is what we get. Now, this results in a equivalent magnetic field let me call it b final that is mu 0 j z. So, which is mu 0 lambda omega over 2 pi along the unit vector z. Now, since the angular momentum is in the same direction I can borrow this expression. So, if I know that the induced final magnetic field happens to be b f then I should be able to write down what is the final angular momentum of the field which is b final lambda a square by 2 along the z direction and if you plug the value of b final which we just now have calculated then this works what I have gives mu 0 by 4 pi lambda square a square omega. So, this is my final angular momentum of the electromagnetic field you notice that the initial angular momentum is not the same as this. So, in order to compensate for this change in the angular momentum my system let us assume that the moment of inertia of that system is i then I will have i omega plus this quantity there mu 0 by 4 pi lambda square a square omega this quantity must be my l i which is b 0 lambda square a square by 2. So, this is this is what I get b 0 lambda a square by 2 and this allows us to determine what should be the value of the what should be the value of the angular velocity with which the disk will rotate. The summary of this is that in considering conservation of angular momentum of a system which has sources and sources of electromagnetic field one should take into account the total contribution both the contribution to angular momentum from the sources as well as due to the electromagnetic field. The next application that I talk about is that we have seen that the electromagnetic field carries momentum. So, as a result supposing I put electromagnetic field in an enclosure a cavity then since this carries momentum the walls of that cavity will experience pressure in the presence of electromagnetic field and this actually is something which you can sort of verify experimentally when electromagnetic waves falls on let us say a mirror. So, the way we will calculate this is this that the force supposing I have a cavity and so this is my cavity inside there is electromagnetic field and let us suppose I am talking about this wall which has this as the direction of its normal. So, the electromagnetic waves or electromagnetic field will exert a pressure on this right hand wall and I know that this force supposing I look at a small area d s on this wall then the force on that wall will be given by the Maxwell stress tensor dotted with d s. Now, since my force is in the x direction the direction of d s is in the x direction all that I require is just the x x component of the Maxwell stress tensor and that we have seen is epsilon 0 e x square minus half e square plus 1 over mu 0 b x square minus half b square. Now, if my radiation is isotropic that is exerting identical forces in all directions then it makes sense to say that e x square is actually a third of e square and likewise b x square is one third of b square. So, we plug this in what you get is d f of course, only the x component is one sixth epsilon 0 e square plus one sixth b square over mu 0 and if you look at this expression you notice this is one third of half epsilon 0 e square plus half b square over mu 0 which is nothing, but the energy this is nothing, but the energy density of the electromagnetic field. So, this is energy density divided by 3 we will talk about some applications of this incidentally this is the starting point for derivation of something like a Stefan Boltzmann law. With this I have completed the discussion of various conservation laws which arise or which have to be taken into account when we discuss electromagnetic fields. What I am going to do now is to go over to the set of equations that we wrote down earlier and obtain what are known as plane wave solutions to the electromagnetic field equations. The electromagnetic field equations are a set of equations which have some special solutions and one of those special solutions is the solution of the plane wave solution we will be talking about that. To begin with I will be taking talking about a linear isopropic medium. Linear implying that relationship between d and e is linear the constant of proportionality is the dielectric permittivity and the relationship between b and the h field are also linear. So, b is equal to mu h I will also talk about regions which are source free that is the solutions that I am going to look for are in regions where rho is equal to 0 and j is equal to 0. Now, that tells me that in the source free region del dot of e must be equal to 0 because there is no rho and of course del dot of b is always equal to 0. Now, del cross e which is the Faraday's law is equal to minus d b by d t and del cross b I had a mu 0 of j which I will drop because my j is equal to 0. So, this is equal to mu epsilon let me take mu epsilon rather than mu 0 epsilon 0 if I had mu 0 epsilon 0 it would mean I am taking the solution in vacuum, but this is mu epsilon d e by d t. So, these are the basic equations with which we will work. Now, notice take any one of these equations for example, take del cross e. Now, if you take del cross of this equation del cross del cross e I get minus d by d t of del cross b and of course we have seen several times this gives me del of del dot of e minus del square e that is equal to minus d by d t of for del cross b I will replace from Maxwell's Ampere's law mu epsilon d e by d t. We had seen that in the source free region del dot of e is equal to 0. So, I get a del square e is mu epsilon d square e by d t square. So, let me write that down. Identically you could have converted this into an equation for the b field and all that you need to do then is to take the del cross b equation, take del cross of this replace for del cross of e from this equation. So, I would get very similar equation del square of b is mu epsilon d square b by d t square. Now, these are equations which sort of look very simple, but there is a problem. See when you take del square of e it does not mean these are actually six equations because e and b are vectors. So, I have to take the component wise, but del square of e if you take x component is not in general equal to del square of e x. This however is true if I work in Cartesian coordinate system. If you were to solve this in for instance the spherical polar then of course, it will mix up the various components. We had seen that del square mixes up, but Cartesian coordinate system you of course, have del square e x del square of e is d square over d x square e x e x plus d square over d y square e y plus d square over d z square e z. So, as a result solving these in Cartesian coordinate system is fairly simple and that is what we are going to do for the moment. Now, what is it that we get now? Look firstly what we have seen is this we have got del dot of e is equal to 0 and del dot of b is equal to 0. This will let me I will return back to this picture in a second. So, I am going to be looking for some special solutions of this equation and these are the what are known as the plane wave solution. What are plane waves? The plane waves are those for which the surfaces of constant phase which are also known as the wave fronts their planes. So, for example, a spherical wave would mean that if you are looking at finding the locus of all points which have the same phase then in a spherical wave those surfaces are spheres, but in plane wave the surfaces of constant phases are planes. So, the solution I am looking for is this. I am looking at e vector as e 0 e to the power i k dot r minus omega t. Now, this is very useful because what we can always do is to of course, our waves are real. So, what we could do is to take an exponential form and ultimately take for example, either real part or imaginary part. If I take a real part I can for example, if e 0 is taken as real then I will get cosine k dot r minus omega t, but otherwise this is the way the algebra becomes very simple if you work with this. Now, and similarly b is equal to b 0 e to the power i k dot r minus omega t. So, let us look at what is meant by then surfaces of constant phase. So, surfaces of constant phase will be those surfaces for which k dot r minus omega t is constant. So, I am looking at incidentally I should point out that I can also have a plus sign there. The difference is in the former case when k dot r minus omega t is there it is what is known as a forward moving wave and if k dot r plus omega t is there it is a backward moving wave. So, the surfaces of constant phase will be k dot r well let me take a minus omega t that is equal to constant. So, let us assume that at any given time omega t is constant. So, this implies that k dot r is constant. Now, let us say the component of r vector along the direction of k, k is known as a propagation vector is zeta let us say. So, therefore, this zeta which is defined as k dot r this is constant. So, therefore, what I have is minus omega t plus k zeta that is equal to constant and if you differentiate it now with respect to time what you get is at a sorry this is this is a this is a constant and. So, I have got k dot r. So, that is k zeta and this quantity is constant. So, therefore, I get minus omega if I differentiate it with respect to time I get k d zeta by d t that is equal to omega which means d zeta by d t is omega by k plus or minus if you had chosen the other sign and which is equal to plus or minus the velocity. So, omega by k is the is what is known as the phase velocity we will see why it is not just velocity, but the phase velocity later. So, these are the harmonic solutions the plane wave solutions of this equation. So, let us look at the plane wave solution little more carefully. So, we had already seen that del dot of e is equal to 0. Now, since I have got e is e to the power i k dot r minus omega t if you took a divergence this will lead to k dot of e is equal to 0. Parallel since del dot of b is equal to 0 I will get k dot b is equal to 0 which tells me remember that we are in a non conducting medium which tells me that electric and the magnetic field vectors are transverse to the propagation vector. That is not all there is something more that happens let us take the one of these equation let us say del cross b equation I know that del cross b is mu epsilon d e by d t. Now, this will result in I k cross b operation of del is equivalent to multiplying with I k because of the exponential form that we have taken. So, that is equal to now operation. So, therefore, let me write this the del operator is same as multiplication by I k and d by d t operator is same as multiplication by minus I omega because I am looking only at forward moving wave. So, this is minus I omega mu epsilon e. Now, what you do is this. So, I and I will go away multiply the take the cross product of both sides with k for instance. Find out what is k cross k cross b and that is equal to well there is a minus sign I have minus omega mu epsilon and then k cross e. This is k cross k cross b and I know which can be written as k times k dot b this left hand side is k times k dot b minus b times k dot k which is k square that is equal to minus omega mu epsilon k cross e k dot b is equal to 0 because the magnetic field is transverse to k. So, this term will go away minus will take care of that. So, therefore, the magnetic field b is omega mu epsilon by k k cross e. So, you could rewrite this as omega mu epsilon unit vector k cross e. Let us examine this mu times epsilon mu times epsilon we have seen is 1 over velocity square. This is seen from this equation directly because this is then a wave equation whose velocity is 1 over mu epsilon. So, therefore, this quantity is written as this is omega I should have had a k square there. So, therefore, there is a 1 over k still outstanding there. So, I have a omega by k and mu epsilon is 1 over v square k cross e and omega by k gives be 1 velocity. So, therefore, this is 1 over v k cross e and if I am doing this thing in vacuum then this velocity is of course, just the c namely the velocity of light. But let us see what I have got so far. What I have got is k dot e is equal to 0 k dot b is equal to 0 b is equal to 1 over velocity times k cross e. What is the import of these equations? These tells me that e and b are perpendicular to k and this tells me that e and b themselves are perpendicular to each other. So, it means that the electric field the magnetic field and the propagation vector k they form a triad a right handed triad they are mutually orthogonal and they form a right handed triad. I will take you back to this picture that I had shown you earlier. So, what it means is this the since the electric field magnetic field and the direction of propagation are perpendicular to each other. The suppose I take the direction of propagation as the z direction then electric and the magnetic field will lie in x y plane. In that plane the electric field and the magnetic field will be perpendicular to each other. So, what it means is this that if I have a propagation like this. Suppose the I have a electric field this way this is magnetic field and this is my propagation vector and as the wave progresses this one moves along k. So, that e and b always remain in one plane and what you see there is just a pictorial representation since e is varying with time sinusoidally this is the green picture perpendicular to that the b is also varying with time one section of this is missing, but that is what the whole thing is about. The let me also then talk about a few other quantities associated with the electromagnetic field electromagnetic waves. For example, we had shown that the energy density is given by an expression like this half epsilon 0 e square plus b square by 2 mu 0 and we have all we have already seen that b is equal to 1 over v k cross e. So, I can simply rewrite this that supposing I want to write down what is energy density. So, I have got half well there is epsilon 0 e square, but I have been a medium. So, let me just generalize half epsilon e square plus b square by 2 mu 0 2 mu and if I take half epsilon out for instance I will get e square plus b square by 2 mu b square by mu epsilon mu epsilon is 1 over mu epsilon is v square. So, 1 by 2 epsilon e square plus v square b square. So, this is nothing, but e square itself. So, therefore, this is epsilon times e square. So, this for example, is one of the ways in which you can write down the total energy density. The .pointing vector s or let us take its magnitude is e cross h I would digress a little bit and sort of you must have noticed that whenever we write pointing vector we prefer to write it as e cross h and not as e cross b though occasionally that would be true if b and h are linear, but a general expression is e cross h. The reason is that if you recall the h field is the field due to real sources and the difference between the b field and the h field for instance could arise because of magnetization and the magnetization currents being bound currents they cannot transport energy. So, therefore, e cross h is the pointing vector and this is nothing, but well I am talking about linear medium. So, it is e b by mu 0 you could also write it as for example, c epsilon 0 e square. If you look at the pointing vector itself this will be then c epsilon 0 e 0 square supposing I am looking at the real part of that exponential then cos square k z minus omega t along the direction k. The intensity which is defined as the time average of the pointing vector I have to then take an average of this over a time period which works out to half. So, this is equal to half c epsilon 0 e 0 square k. So, this is the intensity. Now, before I proceed further I would like to make a comment on what is known as the state of polarization. This is something which is already familiar to you, but let me make some comments in any case. We have said let me for the moment talk only about electric vector because the electric vector being transverse to the magnetic vector. So, if I talk about electric vector you can make similar conclusion about the magnetic vector as well. So, what we have said is that the electric vector lies in a plane perpendicular to the direction of propagation. Now, if that is true. So, this is my direction of propagation and let us say that this is a plane which is perpendicular to it the electric vector can be in any direction thereof. Now, so we had written that since electric vector is perpendicular to the z, z direction. I can express a general electric vector e which depends upon z and t by let us say the real part of. Now, since this is in x y plane I can write it as for example, e 0 x times i plus e 0 y times j and of course, the exponential factor e to the power i omega t minus well minus i omega t plus i k z. Now, in general these quantities e 0 x and e 0 y they are complex. Now, let us take some special relationship. So, suppose my e 0 x supposing is equal to e 0 x magnitude times e to the power i phi. Now, let me also say that e 0 y is also equal to its magnitude and suppose I have the same phase. Suppose, let it have the same phase then my electric field will be written as well real part of. So, e 0 x magnitude i plus e 0 y magnitude j times cosine of k z minus omega t plus phi. Notice this that in this case the magnitude of the vector changes from 0 because cosine can take 0 value to of course, e 0 x square plus e 0 y square, but its direction remains constant. So, the electric field vector as it propagates points in the same direction. Such a thing is what we are calling as linearly polarized wave. Now, what we could do is to take different values. For example, supposing there is a phase difference between the x component and the y component supposing this is e 0 y e to the power i phi. Now, this will then be in general an elliptically polarized wave. Two amplitudes are equal and the phase difference happens to be pi by 2. If phi is equal to pi by 2 and e 0 x is equal to e 0 y I get what is known as a circularly polarized wave. Next lecture we will be talking about how to obtain from electromagnetic theory standard relations like for example, the reflection and refraction of electromagnetic waves.