 Last lecture, we discussed about the fact that an electromagnetic field stores energy and we obtained an expression for the energy density of the electromagnetic field. And what we found is that if you are looking for a system which contains charges and currents, then when you talk about the change in the energy, you have to not only worry about the energy of the sources, you also have to worry about the energy that is stored in the electromagnetic field. And if you have a closed volume, then the any change in the energy of the system is due to the fact that certain amount of energy could be flowing out through the closed surface of the volume. What we want to do today is to prove that just as the electromagnetic field is a storehouse of energy, there is a momentum also associated with electromagnetic field and as well as an angular momentum which is also associated with the electromagnetic field. As a result, if I am looking for a closed system without any external force, then the change I must have conservation of momentum and angular momentum and in applying these rules of conservation of momentum and angular momentum, I need to worry about the changes in the angular momentum and momentum stored in the electromagnetic field itself. So we will, we started talking about it last time and so we will be continuing with it today. But this is simply to tell you that this is going to be there in every lecture that is the set of equations which govern the electrodynamics. This is complete set of equations which are known as the Maxwell's equation. We supplement them with couple of constitutive relation between the polarization electric field and the displacement vector d and similarly between the magnetic field h, the flux density b and the magnetization m. We will be using them regularly and so therefore, almost on every lecture we need to remind this. .. So, let us talk about linear momentum. We start with very basic definition of linear momentum which is the force which is the rate of change of momentum. So, mechanical momentum if you like d p by d t. Now, I know that the, if I have a collection of charges and currents, then this quantity is in the continuum limit is the force due to electric field which is rho e plus the force Lorentz force due to magnetic field which is rho v cross b and of course, d cube x. So, what I do now is that we introduce instead of the e and b the corresponding expressions from the Maxwell's equations that we wrote down. For instance we know that del dot of e is rho by epsilon 0 and del cross b is mu 0 j. So, in this case rho times v, rho times v is the current density. So, therefore, I first express this rho in the first equation as epsilon 0 times del dot e. So, epsilon 0 times I already have an e. So, e times del dot e. Then I have rho v which is j and that is 1 over mu 0. Well, this is actually j cross b. So, therefore, I would write this in terms of the del cross b that I have got. So, this quantity is j cross b and we had seen that in the full Maxwell's equation we had here. So, you notice that let me I am using b and h interchangeably for the simple reason I have assumed b is equal to mu 0 h I am working in free space. So, del cross b is mu 0 j plus mu 0 epsilon 0 d e by d t. And this comes because if you refer to this set of equations you find that del cross h is j free plus d d by d t and once again I am assuming d is equal to epsilon 0 e. So, therefore, I write everything in terms of b and electric field of course I could also work with h, but my system is linear. So, therefore, this mu 0 j that is there I write it as j is equal to 1 over mu 0 del cross b minus epsilon 0 d e by d t. So, that is my f and so let me write it fully. So, f is given by d p mech by d t is equal to epsilon 0 integrals are all over volume e times del dot e d q x plus j cross b which I write as 1 over mu 0 integral over volume del cross b minus mu 0 epsilon 0 which actually happens to be 1 over c square at some stage I will write that also d e by d t. So, this is my expression for the rate of change of total momentum and what I did is to use some algebra and so at this moment I am not disturbing the first term, but in the second term I notice that what I required is another cross e because I had j cross b and what I have written down is just j. So, there is a cross b there. Now, so what I do is this there is this term here d e by d t cross b. So, I have d e by d t cross b which I write it as d by d t of e cross b minus e cross d b by d t and if you recall that from Faraday's law the first term of course I do not change minus d b by d t is del cross of e. So, therefore this term becomes e cross del cross e this term I will replace in place of this. So, if I do that, but before that I will do something else if you look at this expression I have here 1 over mu 0 and mu 0 epsilon 0 d e by d t. So, this I have replaced by this and this. So, therefore 1 mu 0 cancels out I am left with epsilon 0. So, therefore this term when it comes to the other side will just have an epsilon 0. So, let us write it down. So, I get d p mech by d t and I will bring that d by d t of e cross term b term there and since this is an integral the only variation is with respect to total time. So, therefore this is e cross b d cube x recall that there was a minus sign there it has come to that side. So, this quantity is equal to epsilon 0 the electric field term which is e times del dot e d cube x and a few terms which we just now derived namely 1 over mu 0 integral del cross b cross b d cube x minus epsilon 0 integral del cross e cross e, but I have changed the order. So, let me change the sign as well d cube x. So, what I will do is this I want to write this in a little symmetrically you notice that electric field has an additional term there del dot e e times del dot e. The other two terms here are very similar del cross b cross b and del cross e cross e. However, I have a great advantage that del dot of b is equal to 0. So, I can write these symmetrically by putting in at del dot b term. So, let me write this now. So, epsilon 0 integral e times del dot e minus e cross del cross e d cube x and the corresponding magnetic field term which is very symmetric and that is written as b times del dot b which is the term which we have just now added and that is equal to 0 minus b cross del cross b. Rather looks horrible expression, but let us try to sort of see what have we got. On the left hand side I have got d p mech by d t plus epsilon 0 d by d t of e cross b d cube x and we had seen that epsilon 0 e cross b let me recall for you the expression for the energy density. So, we had seen that epsilon 0 e cross b is since my system is linear b is equal to mu 0 h. So, it is epsilon 0 mu 0 e cross h and we defined e cross h as the pointing vector last time and epsilon 0 mu 0 is 1 over c square. So, this is 1 over c square e cross h which is equal to s. So, therefore, my expression on the left hand side I will simply write it because the other one is rather detailed. So, this plus 1 over c square integral s d cube x and that is equal to that expression that we wrote down just little while back a symmetric expression containing between electric field and the magnetic field containing a large number of terms. Now, notice we had said that s is the energy flux. So, let us look at a few things here if s is the energy flux then s by c is the energy density that will be carried by a traveling electromagnetic field and as a result s by c square. So, this is the energy density carried by the electromagnetic field and s by c square happens to be the momentum density because I am dividing it by another velocity. Now, interestingly all our derivations have so far been non-relativistic, but the velocity of light with which we will later on see the electromagnetic waves propagate in free space is coming in very naturally into the problem. So, next question is this that. So, left hand side what I have got is rate of change of this of course, since s by c square is the momentum density this represents the total momentum associated with the electromagnetic field. So, here I have got the mechanical momentum and this is my total momentum. Now, if you remember that when we had this type of a situation that we have a rate of change of momentum and here on the right hand side I should be having a force and so normally we expect the right hand side to be expressed as a gradient of something like a potential which is my force, but notice that expressing this complicated expression as a gradient of a given quantity seems to be rather difficult and it is this quantity which gives us a slightly different way of looking at it and this is what I will be talking about now, but what I will do is this I will express only because this is these two expressions are identical with respect to electric field and the magnetic field other than one has an epsilon 0 the other one has a 1 over mu 0. I will just do the algebra for the first one of the terms let us say the electric field terms and I will simply substitute into a similar expression for the magnetic field term. So, let me write this down that what is it that we have left hand side we already know that we have got total rate of change of total momentum and so therefore, the quantity that I want to now write is density because I am not putting in there integrals. So, E times del dot E minus E cross del cross E, so let us just look at what is the you know this is obviously I expect this whole thing to look like a vector because the left hand side is a vector. So, since I cannot immediately find a simple way of doing it let me just try to write down what is the x component of this quantity. So, here I have got E x del dot E is a scalar. So, since del dot E is a scalar, so I simply write d E x by d x plus d E y by d y plus d E z by d z minus E cross del cross is x component. So, which is E y del cross is z component minus minus plus E z del cross is y component, retain the first term the way it is E x d E x by d x plus d E y by d y plus d E z by d z and expand the two del cross E that I have got which is minus E y into del cross y z. So, it is d by d x of E y minus d by d y of E x the plus E z times this is del cross y y component. So, it is d by d z of E x minus d by d x of E z. So, there are several terms there I want to now sort of try to simplify this a little bit. . First thing that you notice is E x d E x by d x can be written as a half of d by d x of E x square I have a term here E x d E y by d y and a term another minus minus plus d by d y d by d y of E x. So, if you combine this term and that term what I get is I can write that as d by d y of E x E y and identically I have a term here E x d E z by let me just put a double click on it E x d E z by d z and E z E by d z of E x. So, therefore, these two terms will give me plus d by d z of E x E z. What am I left with I am left with these two terms only both of them have negative sign in front of it and this is E y d by d x of E y. So, therefore, this is half both the terms are d by d x term and I will write this as E y square plus E z square. You look at this expression now you notice that in this term if I added an E x square then I will get E x square plus E y square plus E z square which is E square. But if I added an E x square inside the bracket which is like subtracting half of d by d x of E x square. So, I must add another half. So, which will mean this half will go. So, if you now combine all these expressions what you are getting is because all these terms are d by d x this is d by d x this is d by d y this is d by d z. So, what I am getting is that half has gone away. So, I am getting d by d x of E x square plus d by d y of E x E y plus d by d z of E x E z and minus half d by d x of E square. And this if you recall is just the first term that is the x component this is what I am trying to do. So, therefore, when I add up the various components I will get d by d y of E y square d by d x etcetera. How do I simplify this now the thing is. So, I have similar terms two more terms which I must add up now what was found which I will prove by first assuming the result is that the this quantity after I add up the corresponding y and the z components it can be written not in terms of see we had expected it will be a vector and it is a gradient of a scalar quantity. So, you notice that when I expected this to be gradient of a scalar that is going up scalar is one quantity when you take the gradient it becomes a vector which is characterized by three quantities. Now, just as you define a scalar as a quantity having essentially a single quantity a vector is characterized by three quantities in the Cartesian x y z. Now, one can define a quantity which is known as a tensor that a tensor can be of arbitrary rank. So, for instance a tensor of rank two is characterized by nine quantities. So, for example, if I talk about a tensor t just as a vector v is characterized by v x v y v z a tensor t is characterized by a pair of indices. For example, the components of a tensors will be t i j where i going from 1 to 3 1 to 3 and j of course, also going from 1 to 3 can in principle define a tensor of this is tensor of rank two. One can define a tensor of rank three with a quantity characterized by three indices namely t i j k that is of course, 27 quantities. So, this is 3 square quantities. Now, notice one thing that the reason why I cannot express this quantity as a gradient of a potential is because this seems to mix up the components. One good thing is it never mixes up all the components it mixes up maximum two components at a time. Now, that tells us that may be we should not be looking for expressing it as a gradient of a scalar, but we should probably go higher up take talk about a tensor of rank two and of course, by doing an appropriate algebra reduce the tensor of rank two to a vector because after all left hand side is a vector and that is what we do here. So, what one can show which I will show after assuming the result that if you define a tensor of rank two which I will indicate by a notation like this a double arrow such that the alpha beta components alpha going from 1 to 3 is epsilon 0 e alpha e beta minus half of e square times delta alpha beta. Well remember that I worked only on the electric field. So, what I have is actually a corresponding term from the magnetic field as well and this will be b alpha b beta minus half b square delta alpha beta. This quantity has been given a name as Maxwell's stress tensor. What you want to do now is to obtain the relationship between what I proved here that is the remember the integration of this just x component add up the corresponding y and the z component take the integration then you should be able to show that this is nothing but the momentum that appears on the left hand side. So, let us let us look at how one handles this. So, basically this is what we have written down, but let me let me illustrate it by sort of writing down specifically some components. For example, let me take b is equal to 0 as an example. Now, if you take b is equal to 0 my stress tensor only has electric component. So, the t can be then expressed as a matrix which is epsilon 0. Now, remember this is x x x y x z like this it goes. So, x x is E x square because E x E x minus half E square because it is x x. Now, this is E x E y the delta alpha beta term does not coming. This is E x E z this is a symmetric tensor because 1 2 2 1. So, this is also though E y E x I will write it as E x E y this is E y square minus half E square this is E y E z and this is E x E z this is E y E z and this is E z square minus half E square. So, this is this is the way one could write down. So, this is t x x this is t x y this is t x z t y x t y y y t y z etcetera. And if you do not have magnetic field equal to 0 then of course, this will become much bigger. Now, what we want to show is this that if you take a divergence of this quantity then you get hold of a scale vector. Now, remember that when we had a vector if you did a divergence you got a scalar. In other words the a vector is a tensor of rank 1. So, when you took a divergence of a vector which is divergence of a tensor of rank 1 you got a scalar which can be regarded as a tensor of rank 0. So, if I take divergence of a tensor of rank 2 I expect a vector and that is what I am looking for. So, let us look at that. So, del dot t t is the tensor this quantity is for example, if I look at the alpha component of this it is a vector. So, there are three components there. So, this is sum over beta is equal to 1 2 3 d by d x beta of t alpha beta this beta index is summed and therefore, it is just that. So, let us look take again b is equal to 0 because it is a symmetric term. So, therefore, whatever I do for e I can carry it over for b. So, what I will show is that if you take del dot t and let us look at its x component and let me take b equal to 0 for convenience. So, by definition here is the definition therefore, I get d by d x of t x x plus d by d y of t x y plus d by d z of t x z. Now, remember I had already derived what is t x x what is t x y what is t x z. So, this is equal to d by d x of e x square minus half e square plus d by d y of t x y is e x e y plus d by d z of t x z is e x e z. Now, this is precisely what we had shown to be the x component of the quantities which are there. So, you notice this that I wrote down this is the way I had written it down this is exactly the x component. So, in other words the Maxwell stress tensor that we have defined its divergence which makes it a vector gives me the right hand side of the equation that I wrote down. So, let us now repeat rewrite the equations they are summarized on the screen. So, this is d p mech by d t plus we had seen this is 1 over c square d s by d t. So, that is equal to divergence of the Maxwell stress tensor. So, if you now want to write down if you remember that this is my way I wrote down, but this is total d p mech by d t this should have been integrated out because that was a density. So, this side should also be integrated out and as a result one can prove a theorem very similar to the way we prove the divergence theorem. So, this quantity will give you the surface integral of t dot d s exactly the way we did it t dot m can be interpreted as the momentum flux which is normal to the boundary surface. Now if instead of this total momentum total electromagnetic field momentum you want to write it in terms of the momentum density then what you get is d p mech by d t small p I am using for density plus 1 over c square d s by d t is equal to del dot. The interpretation of this is very similar to the way we interpreted the energy density. So, I have got on the left hand side the net electromagnetic momentum and the mechanical momentum and any change in this can happen due to the momentum flux that is going out of the surface. So, in some sense this equation is an equation for conservation of linear momentum. Give some very simple illustrations there of this one is a rather simple example I will give a little more complicated example later. Just consider a pair of capacitor plates we know that let me take as the positive x direction this is the positively charged plate this is the negatively charged plate. So, the electric field goes from the positive charged plate to the negative charged plate inside the between the capacitor plates. . So, if you look back and look at what is the field the Maxwell's stress tensor is like the you notice that the stress tensor is E x square minus E square. Now remember electric field is in the same direction as the x direction. So, therefore, E x square and E square are the same which means this term should be E square by 2 and since there are only x components any cross terms will be 0. So, this tensor will be a diagonal matrix and here what I have written down is the strength of the electric field which is sigma square by 2 epsilon 0 1 minus 1 minus 1. So, that is this should have been squared and the if you look at f equal to t dot t dot d s then you can simply find out how much is the pressure on the negative plate. Let me take a rather difficult example and this is a problem which we had talked about earlier when we talked about the way to discuss the magnetostatics, but this time I will I am looking at it as an illustration of how to use Maxwell's stress tensor. So, this is I have a shell which is a charged cell and the charge is charge density is sigma and this is rotating with an angular velocity omega. What we want to do is to find out how much is the force exerted by the southern hemisphere of this charge sphere on the northern hemisphere. So, it is a rotating charge disk. Now, you in the during when we did magnetostatics we had taken up this type of a problem and we had seen that here is the coordinate system. This is your radial vector perpendicular to that along the direction of increasing theta is the direction of theta vector and this is the z. This is the standard cylindrical coordinate system if you like. Now, the magnetic field inside was constant magnetic field inside was given by two-third mu 0 sigma r omega along the z direction. Magnetic field outside had this expression mu 0 m by 4 pi r cube where m is a magnetic moment and which is related to this by 4 pi by 3 m actually it is r cube into r omega sigma, but r 4th omega sigma and this is the expression for the magnetic field outside. . So, what we are going to do is to try to find write it in terms of the Maxwell's stress tensor. Let me let me emphasize that this is not the simplest way of doing this problem, but nevertheless we are looking at it as an application of the Maxwell's stress tensor. So, let us look at what we have. So, firstly you have to realize that by symmetry the force on the northern hemisphere must be along the z direction. Now, since it is along the z direction I should be looking for only f z. So, if I look at f z we have seen f z can be written as the z component of del dot t which has t x y t x z and things like that. So, let us we are not interested in all the t axis. So, we are interested in t one of the components is fixed as z. So, t z x remember there is another point that so far I was singling out the electric field. Now, in this case I am only looking at the magnetic force. So, as a result I will assume that the electric field is 0. Now, this is not really true because the electric field will be always there the electric force will be always there because there is a charge density there. So, therefore, but that is a different part whether it is spinning or not there will be electric force I am not looking at that. So, t z x is 1 over mu 0 b z b x t z y is 1 over mu 0 b z b y and the diagonal component which was t z z is 1 over mu 0 remember this was b z square minus half b square. And how much is f z this we had seen f z was the surface integral of t dot d s t dot d s is a vector though there is a dot, but you must realize this is a dot product with a tensor. So, therefore, it is z component which is equal to z x d s x plus t z y d s y plus t z z d s z. Now, let us look at how we can write this t z x is b z b x. So, this is b z b x d s x. So, I get a other than b z I get a b x d s x from this term I get a b z b y. So, I get b z into b y d s y. So, therefore, I can get a 1 over mu 0 I have forgotten. So, let me know 1 over mu 0 will come here 1 over mu 0. So, what we have seen is from these 3 terms I will get a b z and a b x d s x plus b y d s y plus b z d s z which is nothing, but b dot d s t z z you remember had an additional terms which is. So, I must write it as minus 1 over 2 mu 0 integral b square this term d s z only. So, this is my f z. Now, I am going to calculate both of them separately. Now, so in all words I will have to calculate a contribution due to the internal field which you have seen is constant and an external field. I will illustrate just one of those things. So, look at what is f z for the fields inside. So, when you are looking at the inside the cap that is there is the equatorial plane. So, the only thing that I have there is the pi r square and the magnetic field is constant there along the z direction. So, as a result this simply gives me minus t z z and pi r square minus because it is in negative direction and I know what is t z z. So, which is 1 over mu 0 b z square minus b square by 2 into pi r square and we have seen that inside b square and b z squares are the same. So, therefore this takes care of minus 1 over 2 mu 0 b z square or b square whatever into pi r square. Substitute the constant field which we have here which is two thirds mu 0 omega sigma r whole square and you can simply write this as minus 2 by 9 of pi mu 0 omega square sigma square r fourth. So, this is this is my force on the hemisphere northern hemisphere due to the field which is inside. Now, let us look at what is happening outside the outside expression is a little more complicated, but the principle is more or less the same. So, look if you look at it b outside is mu 0 m by 4 pi r cube I have told you m I have given you an expression into 2 cos theta r plus sin theta theta and I need remember my first term was b z times b dot d s. So, I need that and of course, there is another term which is the second term which is d s z. So, what is b z b z is r dot z the r dot z is cos theta I already have a cos theta here. So, that gives me 2 cos square theta and what is theta dot z theta dot z is minus sin theta minus because the angle theta is the direction the angle that the radial vector makes with the z axis. So, that increasing theta direction is gives you a minus sign. So, it is 2 cos square theta minus sin square theta and. So, you can rewrite this as mu 0 m by 4 pi r cube 3 cos square theta minus 1 what is b dot d s well the magnetic field is given by this. Now, d s on the surface of the hemisphere is along the radial direction. So, therefore, I do not have to worry about this term I simply worry about. So, r dot r is 2 cos theta and the surface element is r square sin theta d theta d phi. So, if you plug this in and the secondly I need a b square term which is simply taking the square of that modular square of this which is 4 cos square theta plus sin square theta add them up you find b square is given by this expression times 3 cos square theta plus 1. With this you fix both the b z and b square which are required for the calculation of the stress tensor. So, this is what we have written down. So, let us look at any one of those integrations. So, b z was this b dot d s was this what is the surface integral of b z b dot d s. Now, this is absolutely trivial because all that is you have to realize is these are cos theta integration and there is a sin theta d theta there the phi integration does not come into the picture. Since it is a hemisphere the integral limits are from 0 to pi by 2 you work that out you get substitute for m you get an expression like this. I can repeat that for the second term which is b square d s z once again the integrals are absolutely straight forward because the integrals are on cos square theta and the surface element gives me a sin theta you add this. So, what I need to do now is to add up all the three things that we have got two terms of the field outside and a term that we obtain for the field inside and you get that the force that is acting on the northern hemisphere is given by an expression on this time.