 Hello and welcome to module 4 of Chemical Kinetics and Transition State Theory. So far we have covered the prerequisites of this course. We will move on with some very basic fundamentals. We will today think on how dynamics happens on the most fundamental level on atomic scales. So, we will introduce the density of states today and finally, conclude with how to calculate the density at thermal equilibrium which is called the Boltzmann distribution. So, last module we ended with the most fundamental equation in chemical kinetics which is called the Arrhenius equation. K is equal to A into e to the power of minus E A over RT. We motivated this equation via the Vanthox analysis and ended with analysis by Arrhenius on the physical interpretation of this equation. What we are going to focus from now on throughout this course is how do we calculate this K. So, if a new reaction is proposed to you and experimentally tell you that I want to do this reaction and experimentally says not done the experiment, but wants to know how what will be the rate constant is it feasible or not. So, what theories you can do to calculate this rate constant K. So, to do that we will have to first understand the dynamics on atomistic scale. Well, the dynamics on atomistic scale is governed by what is called the Schrodinger's equation by quantum mechanics. However, we very well know that for massive particles Newton's laws is well applicable. So, what we will approximate in this course specifically is that the dynamics of the nuclei is governed by Newton's laws. So, let us start formally with defining a box of particles which has n particles and what is the formal definition of Newton's second law for that. So, imagine I have a box and it has n particles in it somewhere in this box 1, 2, 3 so on till particle number n. So, for each particle I have an x, y and z. So, my coordinates are given by a lot of positions. So, it is x 1, y 1, z 1, x y z of particle 1, x 2, y 2, z 2 so on till x n, y n, z n. So, can you tell me how many number of variables I have written here for n particles it is 3n. So, I have written 3n coordinates. There are 3 coordinates for each particle x, y and z. So, for n particles I have 3n coordinates. Not only that I also have velocities of these n particles v x 1, v y 1, v z 1, v x 2, v y 2, v z 2 till till v x n, v y n and v z n. So, I have 3n velocities as well. These becomes a lot of variables and writing them becomes a bit complicated. What we do instead is we define the 3n coordinates as simply q 1, q 2 till q 3n and this we call as the vector q. So, I had 3n variables x 1, y 1, z 1 till till x n, y n, z n. I just want to simplify my notation and for that purpose only I start using q 1, q 2, q 3 till q 3n. So, what you can imagine in your head is that q 1 is x 1, q 2 is y 1, q 3 is z 1 and so on and so forth. I will do the same thing for velocities. I will just simply call them as v 1, v 2 till v 3n and call this the vector v. So, what are the Newton's laws of motion? The Newton's laws of motion I am particularly talking of the second law. The first one is by definition that is how velocity is defined del xi over del t is vi and acceleration is defined to be del vi over del t and this is equal to well it is equal to acceleration is force divided by mass. Now, we also know that for all fundamental forces this fi can be given as some function that looks like this. So, we have a set of differential equations, differential of x is v and differential of v is given by some differential of potential function divided by mass. So, v is potential and it is a function of only x and it might in general depend on all x. So, these are called the Newton's second law. It is really the same as f equal to ma is what I have written for n particle system. One thing I just want to bring up for this such a system where potential is defined it turns out there is a constant of motion which is energy. So, I define energy as a sum of half m vi square this is kinetic energy and this is potential energy. What one can show energy is a constant that is del E over del t is equal to 0. Because this is well known I am just highlighting it here for a specific reason we will soon enough start looking at some functions which will look like this. So, the question that we want to address why are we doing this is so that we can relate macroscopic quantities let us say like temperature or pressure with microscopic states which is given by this x and v's and how do we do that? We do that basically what you think of when you think of temperature you have a box with a lot of particles you have an abrogator number of particles. So, you do not want to find position and velocity of each particle that will be I need 10 to the power of 23 positions and 10 to the power of 23 velocities I will go crazy. So, instead we transform to a different language that language is in this box how much is the density of particles at a given position. So, we transform to a different coordinate different way of looking at things. So, to do that what we do is first so just bear with me for a couple of slides and then it should become very clear to you what I am doing. First we transform from q comma v to q comma p where p is momentum. So, I define formally p to be m into v that is one way to define it there are more rigorous definitions. So, what is going on? So, now we have 6 n coordinates we have q 1 q 2 till q 3 n and we also have 3 n momentum. And we are going to think of the dynamics of these 6 n variables as a function of time. So, in effect what I have done mathematically is that my dynamics is happening in a 6 n dimensional space because that is just a little bit of an extra for you just to get a little bit of flavor of what the mathematics is you do not have to necessarily know this for this course. But what we are doing really is we have let us say this is dimension x 1 this is dimension p 1 similarly I have much many more dimensions that I cannot even draw. So, I have x 2 here and I have some fourth dimension p 2 and I have many, many more such dimensions and I have a particle here in this 6 n dimensional space moving around with. So, that is the way to think about my system now in a mathematical language. Do not worry if this is not mathematically clear we will continue on and hopefully this will keep on in grading in. So, spend some time with it and think whether this gets to you or not. So, let me provide you an example in 1D just to understand this a little better. In 1D let us say I have only x 1 and p 1 just so that I am able to draw some figures I do not know how to draw a 6 n dimensional figure nobody can really draw it on a piece of paper. So, what I am just to give you a sense I also think the same way we are trying to build a knowledge you own 1D and we will be able to perhaps extrapolate it to 6 n dimension. So, I have only x 1 and p 1. So, these are my coordinates now. Let us imagine my potential is that of a harmonic oscillator that is how my potential is given for a harmonic oscillator it is half m omega square x 1 square where omega 1 is omega is the frequency of the harmonic oscillator. Well, I actually know the solution of this let us say at x t equal to 0 is some x naught and let us say for just for example, p at t equal to 0 is 0. So, this problem is exactly solvable and you can quickly do the math for yourself and this will come out to be ok. So, you can verify whether this is true or not. What I want to highlight is not really the details of the solution I have some solution ok just take it for now I want to understand how this looks in phase space. So, think of it as follows at t equal to 0 let me just draw a table x 1 and p 1 I had x naught and I had 0 here. So, let me mark this point in phase space p 1 is 0. So, I get a point here I am assuming x naught to be greater than 0 for example. Now, as time progresses some delta t well x naught is going to decrease and momentum is going to increase yes. So, if I plot a simpler plot x looks like this. So, this is x naught. So, you see as time progresses x is decreasing. So, my particle is going to move in this direction, but you see on the same point my momentum will increase. So, I am moving actually in this direction. So, my next point will be somewhere here ok. So, I get a trajectory that will look like this. I continue this along till momentum becomes some positive number and x becomes 0. So, that is this point x will become negative it will reach here it will come back and will take a full ellipse back to where it started. So, you start at x naught comma 0 you make some trajectory and you come back. So, this is for a simple harmonic oscillator for a different potential your trajectory will be different. So, that is what is meant by dynamics in phase space. So, now, we will come to our central question which is the density of particles. I am not interested in finding position and momentum of each particle of an abacadro number of particles. So, I define density of particles which is the density of finding the system at x comma p. So, the idea is that I am in some high dimensional space. I am drawing only 3 dimension because that is what my limitation is it is truly in 6 n dimension though. I have some point here which is x comma p and I basically consider a small box here in 6 n dimensions again and I find what is the probability that my system is found in that little box of dimensions dx into dp. So, I want to before moving forward I want to just highlight some of the properties of phase space. I am not going to prove all of this, but simply take these properties from classical mechanics. So, by the way a little bit of history this was basically developed by Hamilton. He was actually an abstract mathematician and mathematicians live in a world of their own. They do not really live in a real world and he was trying to figure out properties of some equations and he got interested in the differential equation which is the Newton's laws of motion. That is the way he thought about it and he I was able to identify some very key features of Newton's laws which are much more general than how Newton originally had written. So, a few things that we specify. First the q and p are all independent of each other. So, I can vary each variable q or p completely independent of what other variable is doing. Mathematically this means that del qi over del qj is 0. What it means is that del qi over del qj is 0 for all i not equal to j. The same applies for pis and pjs. All the momentums are independent of each other. But what might come as a little bit of a surprise to you is that del qi over del pj is 0 for all ij. So, you this initially if you have never seen this comes as a bit of a surprise because you might think of pi as m qi dot. It looks like pi and qi share some relationship but they really do not. Pi and qi dot share a relationship not pi and qi. So, what I am really saying is at a given position the particles momentum can be anything. If I tell you the position of the particle you still have no knowledge of the momentum of the particle. So, that is what it means that saying that del qi over del pj is 0 for all i and j. The second property that we will look at is an integral over phase space. How do I integrate? If I have to do some I have some let us say variable a as a function of q comma p and I want to integrate this function over all phase space and we are going to do this integral soon enough. Basically what we do is we integrate over each variable 1 by 1 over entire range which is minus infinity to plus infinity. So, in short the short hand notation of this we will use is like this. So, when I write this this big integral is what I mean. So, that is just a short hand notation nothing more. One property of density matrix that I want to really highlight. If I integrate density matrix over all coordinates and momentum what should I get? I should get 1 because remember what is density? Density is this rho is the density of finding the system in a given point x comma p. Well, if I integrate over all x and p the system has to be somewhere. So, the whole integral must come out to be 1. It is important to note it does not come out to be 3n it is comes out to be 1 it is you can think of that as a matter of convention. So, this property is a very important property of density matrices. What we are interested in this course these density of matrices are interesting and there is a lot of research on this on how to understand these density matrices. We are interested in the equilibrium density matrix. Equilibrium density matrix is essentially the density matrix at equilibrium. At equilibrium one very important property is that the density matrix will be independent of time that is what equilibrium really means. So, I will have del rho equilibrium over del t must be equal to 0. So, that is what we are trying to find out that is what today's aim is what is rho equilibrium. So, how do we find this we know the Newton's laws, but we have q comma p is slightly more complex dynamics occurring in the 6 n dimensional phase space. So, how do we make progress? So, we first define what is called the Hamiltonian of the system it is a function this is defined to be the kinetic energy in momentum space where this q is potential. So, there is a little bit of difference of this Hamiltonian with energy Hamiltonian is a function. So, you give me a value of these 6 n coordinates of all these 3 n q's and 3 n p's and I will return back you a number which is this sum over i p i square over 2 m plus potential energy. So, it is simply a definition and as it turns out this Hamiltonian is as you can imagine is a constant of time for Newtonian mechanics. So, what we want to get to is what is called the Hamilton's equations of motion. So, again Hamilton was this abstract mathematician who derived all these. So, Hamilton basically derived that x i dot is equal to del h over del p i and p i dot is minus del h over del x i. So, for the previous Hamiltonian that we had written we are going to verify whether this is correct or not. So, we will insert that and find del over del p i we will just use a different summation index not to confuse ourselves this is equal to sum over k 1 over 2 m del over del p i of p k square plus del over del p i of v of x the first thing to note is that positions are independent of momentum remember that is one of the properties we mentioned. So, this term is 0 all positions are completely independent of momentum. Now, this term basically what I will get is simply 1 over 2 m del over del p i of only p i square all other momentum are also independent of p i. So, at the end I get p i to p i over 2 m which is p i over m which is v i. So, you note you get x i dot equal to v i which is what you expected out of Newton's laws and we are going to consider the other point p i dot as well. So, here we will have minus del over del x i sum over k p k square over 2 m plus v of x. Now, you will notice that the first term will vanish because all momentum are also independent of positions. So, this term vanishes that is derivative with respect to x i vanishes and what I get is del over del x i of v of x. Sorry I forgot the arrow there. But this is exactly what we had defined as Newton's laws yeah. So, p i dot is nothing but m v i dot. So, you get back Newton's laws from these Hamilton's equations of motion. So, in summary in this module number 4 what we have introduced are a few basic concepts. First is the notion of phase space which comprises of positions and momentum that is important to note it is momentum and not velocity. Second we have looked at the density matrix in this phase space. And finally, we have looked at the dynamics in this phase space and the dynamics in this phase space is given by the Hamilton's equation of motion which are summarized here del q i over del t is del h over del p i and del p i over del t is minus del h over del q i. So, we have verified that this equation is true for the given Hamiltonian which is kinetic energy plus potential energy. In the next module we will use these ideas and derive the equilibrium density matrix and use this density equilibrium density matrix to calculate useful properties. Thank you very much.