 So, before we go to the non-ibelian case, I just wanted to highlight some of the salient points of the abelian. So, what we saw so far was that we have electromagnetism. So, assume is that we have which happens to be Coulomb law and Faraday's law together, sorry Ampere's law together right because actually there are sources here. So, in the laws of electromagnetism only the Coulomb law and the Ampere law has a current as a source which generates curl of a magnetic field. So, those are those the other two equations simply become identities ok. These two are the laws Faraday's law which has no source and it remember it only says that rate of change of magnetic flux in a region causes electric current electric field to be set up and that electric field is non-conservative. It is the distinction between the EMF generated in Faraday's law and the Coulomb law electric field is that Coulomb law electric field is single is generated by a source. So, it is what is the word. So, its line integrals are 0 if there is. So, what is the conservative field curl is equal to 0 right. So, the Coulomb law fields obey this, but here of course, it is equal to minus d B by d t. So, the E field that is generated by this is not a conservative like vector field and. So, that law transforms in. So, if you will put the indices. So, I hope everybody knows that there is a mapping from F mu nu language to E B language with E 1 E 2 E 3 here and it is anti symmetric tensor and there is B. So, it is 0 1 2 3 and so, where you have 1 2 it will be B 3 where your 1 3 it is minus B 2 and B 3 B 2 3 is equal to B 1 2 3 1 like this. I think these signs are correct if not then the B sector signs are opposite one has to check once well if you want to quickly check we can check B 3 should be equal to it is curl of A. So, 3 is d 2 A 1 minus d 1 A 2 right and that tellies with it being equal to F 2 1 correct. So, F 2 1 must be B 3 which is correct 2 no I have 1 2 is so, good. So, this is opposite right this is correct is not it B equal to. So, curl a third component should start with 2 1. So, how will we would write i j k and then d x d y d z and then A x A y A z. So, 3 would start with 1 2 good. So, then the sign was correct. So, it is equal to and this is nothing, but d 0 A 1 minus A 1 d 0 etcetera. So, I think that sign is correct. So, these laws reproduce this the main thing I want to emphasize is to remember that the F mu nu is intrinsically anti symmetric. In fact, this E and B split is completely artificial it was convenient for low energy physics because you are only looking for 3 vectors and the E field is certainly a genuine 3 vector, B field is not a genuine 3 vector, but it is a good coincidence of 3 dimensions that the curl has exactly as many components as the usual vector field right. An anti symmetric F i j has as many components as n into n minus 1 by 2 in n dimensions. So, in 3 dimensions it will have exactly 3 into 2 by 2 equal to 3. So, this is a miracle of 3 dimensions. So, the anti symmetric tensor will have exactly same number of components as an ordinary vector and that is why you have all these things that tell you if you rotate like this then this points this way like a hair coming out of somebody's head no it is not like that it is really a anti symmetric field anti symmetric tensor. And so, here it all comes together nicely and they fit together as a anti symmetric derivative of 4 vector field and then you can see that this equation is an identity. If F mu nu is anti symmetric then it is easy to check that regardless of what F mu nu is. So, long as it is it changes sign under exchange of the 2 objects or so, long as it is expressed like this d mu nu you can just check this out all the terms will add to 0. So, it is an identity. So, the great Faraday's law is essentially an identity was obvious and the other law is of course, which has no name which is the divergence b is equal to 0. So, those are automatic identities that follow from the F mu nu being anti symmetric. Now then the question is how do you write the action for such a system and also. So, some there are some remarks which we do not want to go too much into the details of, but one thing is that when you come to the particle interpretation you know that electromagnetism the photons have only 2 orientations 2 polarizations. So, there are only 2 degrees of freedom, but there are 4 degrees of freedom here. So, in the field description. So, the covariant field description for I wrote this last time also, but let me just write it again. Whereas, physical particles have only 2 degrees of freedom. So, there is a redundancy and we can see the redundancy easily because actually we know that it is a spin 1 particle more correctly helicity 1 particle. Now, we know that if it was spin 1 then from in quantum mechanics there will only be we only expect minus 1 0 and 1 projections if it is a spin 1 particle and helicity 1 actually means that it is massless. Helicity is in the massless limit plus or minus 1 the 0 part drops out because that you get the spin less component of the multiplet is available only if you go to its rest frame. So, if you have a massive particle which has spin 1 if you go its rest frame then this after all is a helicity. So, you can get the spin 0 component if you can go to its rest frame, but if you can never go to its rest frame then it will be always spinning and you can never overtake it. So, you cannot convert a plus helicity particle to negative helicity particle, but here you can get it to come to 0 then the helicity will then become 0 p dot sigma right. Now therefore, we observe that out of the four components this is more or less what rotation group needs because it is a massive particle you can be in its rest frame and then rotations really are enough to characterize what are the degrees of freedom in it. So, actually therefore, fourth component A 0 seems superfluous the 3 A i should be enough, but when we go to this then here even only the transverse components matter. Now, there is Helmholtz theorem from vector calculus that says that any vector A can be written as A longitudinal plus A transverse where uniquely written the transverse component has divergent 0 and the longitudinal components curless to be 0. So, there is a further restriction because the longitudinal part is set to 0 to get the final 2 degrees of freedom. So, that is how the connection between the covariant field description and the photon or particle description works. Now because of all this the Lagrangian has a specific form. So, this is 1 then 2 as a result of 1 we. So, the procal Lagrangian 1 puts back m square and then there are 3 degrees of freedom because mass is not equal to 0. So, that is this case and there will be 3 degrees of freedom the A 0 is superfluous. We can see that A 0 is superfluous because for both the Lagrangians what is pi 0 I mean the canonical momentum right canonical momentum 0th component would be d 0 time derivative of variation of L which is time derivative of A 0, but there is no A 0 in this because the no time derivative of A 0 because this is always anti symmetric. Whereas, this is d 0 of A 0 there is no such term going to occur in F mu nu right. So, the Lagrangian density does not contain d 0 A 0 at all and so, this is equal to 0. So, it basically says the canonically conjugate quantity to A 0 is 0 there is really no dynamics in A 0 field and that is what we mean by saying A 0 is superfluous. Another way of seeing it is so, this is an exercise check that the most general or should since it is an exercise which we have to think about I will say quadratic order in derivatives first derivative nothing terribly complicated right. Since, as a Lagrangian its you know it starts with kinetic energy. So, it has square of the time derivative. So, in covariant notation it would involve all the derivatives up to quadratic order. So, that covariant Lagrangian that is of this form and then check the condition for so, make a choice. So, here the parameters will be arbitrary. So, choose such that kinetic energy is correctly normalized for each component and such that pi 0 comes out negative comes out 0. So, it will look a little mysterious, but once you start writing you will know what this is all about. So, you can do this and then you will. So, my claim is that this F mu nu F mu nu if you will expand it out in terms of derivatives of A is quadratic in derivatives of A, but it does not it is not the most general one and you will be led to this particular one by choosing the parameters correctly in it correctly ok. Now, coming back to this part. So, just we saw how A 0 is removed to get radiation we need to also remove the third component which is this longitudinal component which is the nuisance and we can see that the gauge transformation. So, 2 now so, this was the exercise for 2 now remark number 3. So, we note that the Prokha Lagrangian does not have gauge invariance. If you shift A mu by D mu lambda then the F mu nu is unchanged. So, F mu nu F mu nu remains, but if you shift A mu by D mu lambda then this term gets messed up and A mu times D mu lambda kind of terms appear in the Lagrangian whereas, lambda is not something physical. So, you do not want it to appear in other words this term will not respect gauge invariance. But the Maxwell Lagrangian we can always ensure longitudinal part is 0 yeah a l can be set to 0 without affecting a transverse. So, I am sorry these things are look I mean dropping them as remarks, but it is after you have understood what the problem is and if you start thinking about it then these are the answers. The real difficult part is to think of the questions. So, once if you get sufficiently familiar with Maxwell's theory then you begin to ask the questions for which I am giving you the answers. So, you will think about it you will have to go back and forth and think why we are making these statements. So, this is easy to see because A nu equal to A minus gradient sorry we have to say the right, but for the space part this is of course, the transformation. We can see that this never affects the transverse part because we can make sure that it does not affect transverse part by making sure that divergence the grad square lambda is 0 and it does not enter the longitudinal part at all curl of A old, but because curl of grad is 0, but that means in other words since in this decomposition if I take curl of A the curl of L is anyway 0. So, this actually implies that curl of A transverse is equal to transverse right because the longitudinal part does not enter the curl at all. It is an identity to write if you take curl of this on both sides then you get this, but that is automatically 0. So, that equation only affects the transverse parts. Therefore, the gauge transformation only affects the longitudinal parts which actually and therefore, making this degree of freedom superfluous in one way that people impose this. So, now let us just summarize everything. In a sense in textbooks you will see what I am about to say next and if you try to understand why they say that these are the reasons for it. So, number 4 thus photon QFT is most often done with gauge conditions they are called gauge fixing A 0 equal to 0 and divergence A equal to 0. And these both are within the broad class called Lorentz gauge and this Lorentz is not Mr. Henrik Antoon Lorentz, but some cousin who did not write a T in it which is d mu m u equal to 0. What is nice about Lorentz gauge is that it is covariant right d mu m u equal to 0. So, it is setting some scalar piece out of the 4 degrees of freedom to 0, but eventually you need to note that. So, this falls within that right. So, this is the class more general class of gauge this is only one scalar condition. Here we are putting two conditions which also break Lorentz invariance A 0 equal to 0 and divergence A equal to 0 will then satisfy this automatically. So, there are other gauges the other gauges. So, exercise study other gauges for example, A 3 equal to 0. So, these gauge conditions are put because of 1 2 and 3 to take care of these ambiguities and where they arise they can be fixed by usually putting these conditions. So, I have said things somewhat backwards trying to give you the reasoning, but normally books will tell you that these are the gauge conditions we use. So, I think we are ready to go to non-ibelian gauge theory and so to go over to that let us recall. So, recall that in the and I also never said what is a billion and non-ibelian. Recall so far we had d mu of psi equal to d mu plus i g times a mu times on psi as the covariant derivative. I think a g was required here no e g e raised to i g lambda x i of x and a tilde x equal to implied that d mu tilde of psi tilde x was equal to. So, by definition this is equal to d mu plus i g a mu tilde of psi tilde x was equal to e raised to i g lambda of x times good old d mu of psi of x this was the transformation. So, this is the meaning of the word covariant. It transforms the same way as psi itself does. If psi is transformed to e raised to i g lambda times psi then the d mu of psi transforms by the same way. So, the same way as is called covariant. Now, when we got right now the business of ebelian we see that if you carry out successive such transformations then the compound right if I have a lambda 1 then I do lambda 2. So, if we do a tilde just to save space we will stop writing all the detail a tilde equal to a minus d mu of lambda 1 let us say and suppose I make a double tilde equal to a tilde minus d mu of lambda 2 then it is clear that it is equal to a minus d mu of lambda 1 plus lambda 2 and similarly and also as a result psi double tilde will be equal to e raised to i g into bracket lambda 1 plus lambda 2 psi because first transformation e raised to i lambda 1 then you do another one another a lambda 2 just gets added in the exponent. So, this property that psi that the 2 gauge transformations just combined to make a 1 makes the set of all such transformation an ebelian group because you can of course, multiply them in reverse order this answer will remain the same. So, the successive transformations combined to give new transformations that itself is group property number 1 right closure if you do 2 of them you get a new one which is in the same league of things and then you can check associativity existence of inverse and because it is ebelian also that the order does not matter identity of course, is lambda equal to 0. Now we want to think in terms of generalizing to non ebelian case here a first very interesting physical property of matter comes into effect because what do we mean by ebelian non ebelian etcetera well we know there are there is a well known group that can be written in this form this is SU 2 where I have where a u can be written as exponent of i theta by 2 times theta cap dot tau matrices poly matrices or more correctly it should be tau by 2 where theta is the amount of rotation this is the direction of rotations and the generators are tau by 2. So, this is analogous to this i g lambda, but now we want to we want this to act on something you know ultimately the covariant derivative was how the d mu of psi transformed. So, we need a psi on which this u can act, but this u is 2 by 2 matrices. So, right. So, first I just want to motivate that this psi you have to deal with has to be also a at least a 2 component vector it has to be a representation of SU 2. So, the representations of the ebelian group this is u 1 group you know unitary group of unimodular group of determinant well its determinant is a complex number is a magnitude 1 determinant complex number. So, we need psi also to be a representation of SU 2 thus. So, one does take various possible representations higher order ones as well the simplest is the spinor representation. Now, spinor do not start thinking of Dirac equation and all that it just means a 2 component complex valued vector. Historically this already existed.