 did see the video last time right, you remember now. So, given n atoms in a molecule each one has 3 degrees of freedom if you are in 3 dimensional space, 3 n degrees of freedom and then the center of mass of the molecule will remove how many? 3 of the degrees of freedom right which is the you can move the whole molecule from place A to place B, center of mass is what center of mass of the molecule is what is moving and that is not going to change the vibrational degrees of freedom. So, 3 translational degrees of freedom associated the center of mass of the molecule is not playing a role in vibrational motion. Other thing is to rotate the whole molecule if it is linear the rotation about the axis of that linear molecule will not do anything. So, only you will have 2 rotations if it is linear molecule if it is non-linear molecule you have 3 axis of rotation. So, you have to remove the rotational degrees of freedom also from the system. So, totally 3 n minus 6 where n is the number of atoms in a molecule that plays the role for vibrational degrees of freedom if it is non-linear molecule. Is that clear? Atoms 3 n degrees of freedom is total and then 3 degrees of freedom is for translations translational motion 3 for rotational motion. So, 3 n minus 6 is the vibrational degrees of if it is linear, linear molecules. So, I am looking at all these molecules in 3 dimensional space ok. So, linear molecule in 3 dimensions again 3 n to start with this one will be again 3 this one will become 2 degrees of freedom because one about its own axis it does not give you for a linear molecule like carbon monoxide nothing happens. So, the number of degrees of freedom vibrational degrees of freedom for a linear molecule will be 3 n minus 5 degrees of freedom will be the vibrational degrees of freedom. Suppose I tell you that you are in a hypothetical situation that you are going to move only in the 2 dimensional plane then what happens? Suppose I say that 2 d space is where you can move around ok then what all modifications will happen 3 n will become 2 n and so on ok. Translational degrees of freedom will become 2 degrees of freedom. Let me look at non-linear hypothetical molecule ok. How many rotational degrees of freedom? 1. So, totally vibrational degrees of freedom for a molecule with atoms on a plane let us say will be non-linear molecule what I am taking it. So, on a plane, but not lying along a line it is on some equilateral triangle or on a square, but it is just confined to the x y plane or ok. And we have seen various things pictorially or in the video. The relative motions of the atoms without altering the center of mass of the molecule are the ones where we have to also rotation does not add to the rotations. Those are the ones which is going to tell you exactly whether there is stretching, bending which are called breathing modes, asymmetric stretching, symmetric stretching you saw some of these pictures on the video. Can we understand this from your character table, projection operators, reducible representations, tensor product of representation that is the theme of today's lecture. So, assumption is you have not forgotten character table, you know tensor product, you know how to get from the reducible representation the irreducible representations. And then we will see whether we could get the asymmetric stretching, symmetric stretching all these things we would like to understand. On the plane you will have about the point rotation that is all. No, but you have to keep it on the plane you are looking at in three-dimension. Just confine that I am on this floor all the atoms are nothing beyond this floor, no going up or down. But still if I can on the plane like if particle is fixed like one can rotate like this or the other can rotate like this. No those will be the relative motions. The whole molecule you can you cannot rotate relatively different atoms in the molecule that is what you are looking for. You take the whole molecule, rotate the whole molecule, the system is intact. Take the molecule, translate the whole molecule, the system is intact. So, those two are the three, those two are the three degrees of freedom which you have to worry about for translation which is like center of mass of the molecule. Similarly, you have a rotational. See whether this works. Is that clear? Not the relative rotation. All those relative motions are like you know displacements about its mean position which will give you the vibrational degrees of freedom. It will not give you the total rotation of the molecule. If you have seen the video it is just at the axis on which the whole molecule is rotating. Not relative atoms in the molecule. Is that all right? Fine. So, now I am going to try and do a little bit of warm up on classical mechanics to tell you how difficult or how cumbersome it is to solve equations. And then I will lead you to group theory to show how simpler it is. It can work for many complex polyatomic molecules. This is the theme on which I am going to pitch today's lecture. So, this is a warm up which we did last time also for a complex system. Anyway, I have said this. There are a number of degrees of freedom and that degrees of freedom will depend on your number of atoms and all these things have explained on board. So, given that there are s degrees of freedom. So, you just take them x 1 till x s are deviations from the mean positions of the atom. So, it gives you displacements of the atom about its mean position in a molecule. So, those are what we call it as excursions of mass points. And typically you will have a Lagrangian which is dependent on x 1, x 2, x s which I put it as a column vector and row vector in between there will be a matrix. So, that this whole thing will be some kind of a number or is that right? So, you have a matrix, a column vector multiplying the matrix here, x dot is the time derivative of these s degrees of freedom and x dot transpose is the column vector. And then sorry row vector this is a column vector and this is a s cross s matrix ok. And similarly you will have a kinetic potential energy term where it is only on the positions. So, displacements of the mean positions which I call it as excursion. This is the formal Lagrangian which you can write. It will be coupled set of you know if you try to find the equations of motion for this, there will be a coupled set of second order equations to solve right. You might have done this. The only way you can make it into uncoupled equations like this is by doing a suitable diagonalization matrix and then find the basis eta for those diagonal Eigen values. Diagonal Eigen values are the frequencies. I am sure you would have done this. If you have not done it, just take a 2D or a 3D system which is coupled, try to diagonalize the matrix, find the Eigen values then try to use the matrix which allows you to diagonalize such a matrix and that will define for you the position excursions over the etas. Etas are what we call it as a normal modes is that clear? So, with this in mind let us look at an example non-linear molecule. So, I said over all transition molecule is eliminated by imposing that the center of mass of the molecule does not get displaced. So, you put that to be 0 origin. So, you have remember an example like a water molecule right. The two hydrogen atoms and this is the oxygen atom and we have this equation for center of mass of the molecule does not get displaced. So, you can determine what is I have given the bond length in the undisplaced situation as L and the angle made with respect to the oxygen molecule I am calling angle subtended by the two hydrogen on the oxygen molecule is 2 alpha clear. So, location of the center of mass of this molecule can be written down you can find that the x is 0 and y will have this sorry z I am taking this to be in the x and z coordinates. So, now rotation this will clarify your question rotation of the molecule can be eliminated by putting angular momentum about center of mass of molecule to be 0 and if you try to use these two equations you can simplify and write this constraint which has to be satisfied. So, basically what do we see? You have 6 degrees of freedom you have 1, 2 and 3 constraints clear. So, essentially how many degrees of freedom you have with you which will be the vibrational degrees of freedom is 3 degrees of freedom. So, translation and one rotation puts 3 constraints on the 6 degrees of freedom what are the 6 degrees of freedom you have x 2 z 2 capital x capital z x 1 z 1. So, I am looking at it in this the y coordinate is remaining the same. If you take this then you find that there are essentially 3 degrees of vibrations possible and we are interested in looking at the 3 degrees of vibrations. So, I am also trying to show you how systematically you can solve these situations in classical mechanics. So, that you appreciate when I do the group theory that this is doable from group theory. So, for non-linear triatomic molecule. So, we have to have which preserves are the vibrational modes subjected to these 3 constraint. So, the only motion where you are allowed the freedom of displacement or excursions of these atoms in the molecules are either the bond length this length could change this length could independently change and you can also have the angle 2 alpha could also change because of the relative displacements of the atoms in the molecule clear. So, these are the only possibilities which I can pictorially see here which may not be visible when we go to complex polyatomic molecule, but let us understand in the simplest case and map it to group theory. So, that you can believe in group theory for complex polyatomic molecule ok. So, bond length changes this is one possibility which will be 1 vibrational degrees of freedom and then you can also have the bond angle changing ok. These are the 2 possible deformations of the molecule without altering the center of mass of the molecule and rotational degree of freedom ok. So, how does bond length change? You can make this length to change for example, here which I call it as delta L 1 is the change in length of the bond length between capital M and small m for this case and delta L 2 is the change in bond length for the mass with coordinates x 2 z 2 and the capital M with x and z ok. And you can also have a rotational constraint ok. So, in fact, this is given explicitly in Landau-Liftz mechanics book in case somebody is I am only giving the final equations here if anybody is interested to derive these things you should go back and look at Landau-Liftz mechanics ok. So, the rotational deformation delta 2 alpha this is given by these changes in which involves x, x, z all the 6 degrees of freedom and the angle alpha ok. So, so far so good looking at these constraint of center of mass angular momentum being about the center of mass being 0 you can try to rewrite the Lagrangian this looks ad hoc at present, but let us just make a change of variable to a new variable q 1 which is the sum of x 1 plus x 2, q 2 is the difference between x 1 and x 2, q 3 is the sum of z 1 plus z 2 and you will also see that the center of mass of the molecule and the angular momentum about the center of mass being 0 will put in some kind of a condition here between q 1 and z 1 minus z 2. So, I am basically trying to say that if you had two coordinates you could sum up the two coordinates and take the difference they are two linearly independent objects, but when it comes here to z 1 the other one is actually related to q 1 and the angle alpha. So, it is not going to be a new degree of freedom ok, z 1 minus z 2 should not be a new degree of freedom it is going to be constrained. So, q 1, q 2, q 3 are essentially the 3 degrees of freedom which can capture the vibrations of this non-linear molecule made of 3 atoms to find the normal modes or actual vibrational degrees of freedom it could be a linear combination of 1, 2 and 3 that I am not really looked at it right now. I just made a change of variable and then I can write for this system the Lagrangian. So, the Lagrangian will have a potential energy there will be potential energy due to vibrations which involves delta l 1 squared another vibration with delta l 2 squared and that could also be a vibrational mode due to delta of the angle change ok. So, you will have those terms added into it. So, this is what I am saying you will have a spring constant solve simple harmonic. Let us take it to be small displacements let us not do a violent vibration modes are like you know small displacements about its mean positions. So, they will all be simple harmonic and let us take the spring constant due to change in bond lengths to be kappa subscript l and the spring constant due to change in bond angle as another kappa alpha and in terms of q you can write the Lagrangian for the system ok. So, try and redo this rewriting it for the system of 3 masses you have 2 masses and a capital N. So, try to write it out explicitly ok and you will find that you have equations where you have a Lagrangian with q 1 dot squared you have a Lagrangian with q 1 squared sin squared alpha and kappa alpha kappa l is the spring constant for the change in bond length ok. And similarly for q 2 also you will have a piece which depends on kappa l, but you will also have a kappa alpha and so on. So, you have this what do you do next given a Lagrangian you try to find the equations of motion for q 1 q 2 and q 3 this is what you will do right. Everybody is bored looks like everybody knows. So, good if you know it is good if you do not know do not worry, but this is the cumbersome process in which a mechanic students classical mechanic student works out the normal modes in and I am sure they would have got problems in exams and for linear molecule finding the normal modes yes or no yes ok. So, equations of motion for that Lagrangian interestingly you see that this first equation is completely decoupled it has no coupling to q 2 or q 3. So, straight away from here you can read off that q 1 which was defined to be x 1 plus x 2 should be 1 vibrational mode with this frequency and again this equation has a coupling between q 2 and q 3 and this one has between q 3 and q 2. So, they are coupled set of differential equation and you know how to solve it what do you do you write the determinant you try to find a normal modes by diagonalizing such a matrix right. So, I leave it you to check this I will put this slide I want you to check because it will help you to understand the group theory way of doing it right now it is still mechanics whatever I have said. So, you see the steps I have gone through I have gone through systematically, but it is a little tedious you need to literally do these things to figure out that q 1 is a normal mode with coordinate x 1 plus x 2, but what we will see is from group theory we can actually get x 1 plus x 2 and then because we have done this hard work you can appreciate x 1 plus x 2 is a normal mode I get from group theory is that clear. So, some of the steps of math I am not going through right now here because it is not a mechanics course, but anybody is interested you can come and bother me if you are not getting in between steps, but you can look at Landau-Liffschitz and you will be able to get these steps ok. So, I am not doing the diagonalization of this coupled set of two equation to determine what are the normal modes which will be a linear combination of q 2 and q 3 and an orthogonal linear combination involving another mode ok. So, this I am not doing it now and when you diagonalize set you will find two eigenvalues the two eigenvalues are different then only you call it non degenerate right. Do you expect it to be different or do you expect it to be same for this molecule which we are looking at? C 2 v is a symmetry the character table tells you that you should have only 1 D reps. So, you better get all these eigenvalues to be different ok. It has to be different it cannot be degenerate ok. Check it out I am sure you can take this and diagonalize it and find the frequencies ok. So, three vibrational degrees of freedom will involve q 1 q 2 q 3 turns out that q 1 itself is a vibrational mode which is one frequency what is the frequency? Frequency is going to involve this factor times this one is omega square. So, this is non degenerate one normal mode or one vibrational mode which is straight away q 1 q 1 is not talking to q 2 and q 3 linear combination of q 2 and q 3 what linear combination you have to diagonalize this and find out and what are the frequencies ok. I am trying to tell you that you will find each one to have a distinct frequency you cannot have the same frequency. If you are doing other molecules like methane which was shown in the video you will start seeing if you do this rigorous exercise there will be some frequencies which will occur three times because there is a 3 D wrapped in the character or if you had a C 3 v symmetry you will get two times which is associated with the frequency should repeat twice and you should be able to find two distinct normal mode sharing the same frequency. That is why they start doing that mixing and joining together you know you remember and you can see that in the methane molecule there are three degenerate vibrational modes and they combine together and so on. When will you get degenerate comes from the symmetry of the molecule. I hope you are you know appreciating what I am trying to drive in how simple looking how complex looking where group theory can play a very important yeah. You can treat it to be a two dimension or you can also treat it to be a three dimension but then you have to remove the translation along the y direction and the rotations which is involving the other yeah I will come to it. C 2 v in the sense that this is an example of a water molecule and I know that the two I put it as two identical m's if you do a C 2 about the z axis you do see that this hydrogen goes into the other hydrogen and vice versa. It will be still if you treat the z axis to be the C 2 axis it is still true and sigma v to be like the x z plane it is still true ok is that clear. So, I am just looking at it from purely from the action of the group theory also and you can treat it like a 3 D problem but then you have to remove those I have to put the center of mass coordinate for the y also. So, basically I could have given x 1, y 1, z 1 uncomplicated the situation. Instead I thought anyway y is not playing a role with all these symmetries of this molecule I can look at it as a C 2 axis on the z axis which takes the two hydrogen atoms to each other. So, I thought it is better to confine to a simpler 2 D situation but you can extrapolate it to 3 D.