 Welcome back to the next lecture. In the previous lecture we talked about the weight of a configuration. We also discussed that instead of working on w working on log w brings the results more easily. And then we also talked about some constraints which need to be followed because the total number of particles total number of molecules in a system are constant and the total energy is also constant. Also keep in mind that we are so far talking about independent molecules. We have not allowed the molecules to interact so far. So, therefore let us now proceed further. Now our next goal is going to be deriving Boltzmann distribution. And in order to derive the Boltzmann distribution here again we are going to use certain constraints and we are going to discuss these equations that is you remember I was talking about that instead of working on w let us work on d log w. So, we will set up some equations where we will write expressions for d log w and set it equal to 0. And these are going to be constraints these constraints I am going to discuss with you in details now. So, the equations that we are going to derive now is our number 1 d log w to be written in the form of partial derivative. And then we will introduce some constants alpha and beta to transform d log w into another equation where we can set these 2 you know either d n i equal to 0 or this bracketed expression equal to 0 if we set d log w equal to 0. And we are also going to use Lagrange's method of undetermined multipliers and eventually we will come up with this expression. And let us now go step by step in deriving these equations. So, let us start with log w log w instead of w let us work on log w this is a function of n 0 n 1 n 2 etcetera. That means I can write d log w is equal to del log w over del n 0 into d n 0 plus del log w del n 1 d n 1 plus you keep on going right when there is some change you can write the change as d log w. And in terms of partial derivatives on the right hand side these will appear. So, let us now write d log w equal to I will write summation i del log w by del n i into d n i all these upper you know these various parts of the total sum I have summarized into summation i del log w by del n i into d n i. The purpose is to set this d log w equal to 0 we will set this equal to 0. Now suppose that if we set it equal to 0 does that mean that either this del log w by del n i equal to 0 or d n i equal to 0 can we set like that d n i is not equal to 0 because there has to be some change that does that mean that del log w by del n i equal to 0 no the reason is here if you carefully examine the upper steps these d n i's are not independent they are dependent they are appearing with each term. So, therefore, we cannot set either this equal to 0 or this equal to 0 we cannot do that because d n i's are dependent they are not independent. So, therefore, what should be done in such cases so that we can treat d n i's as independent and that is where the method of undetermined multipliers is used, but before I go into the method of undetermined multipliers let us use the constraints let us recall those constraints. So, those constraints first constraint was summation i n i equal to 0 can I write this as for a change summation n d n i equal to 0 that means when you add up the sum of all the changes some are positive some are negative everything total sum of all the changes has to be equal to 0 second one summation i n i e i equal to 0 this was the second constraint. So, therefore, I can write for a change n i or take a derivative n i d e i plus summation i e i d n i this has to be 0 we know that d n i's cannot be 0 energy levels are fixed energy states are fixed therefore, what is 0 this d e i is equal to 0. So, what we have here now is summation i e i d n i equal to 0 this is my constraint number 1 this is my constraint number 2 what is the Lagrange's method of undetermined multipliers is that let us multiply these two constraints with some constant. Let us say I multiply this one with alpha and I multiply this with minus beta why minus beta it will become clear soon see whether you multiply by alpha or beta or minus beta eventually the value is 0 because you are multiplying with 0 only value is 0. And then if I add these two into both if I add into this expression I am going to do that now in the next step that is d log w is equal to summation i del log w by del n i I had d n i plus I have alpha into d n i summation i d n i the value is 0 minus beta into summation i e i d n i is equal to 0 I will set this equal to 0. See I have added from one constraint the other constraint both the values of 0. So, it does not make any difference let us rewrite this again d log w is equal to summation i I have del log w del n i plus alpha minus beta e i and I will take d n i out. Now I will set it equal to 0 remember that these alpha and beta are undetermined multipliers what we are doing is we are introducing some undetermined multipliers such that now we treat these d n i's as independent and then work out for the values of alpha and beta. So, that is the way to deal with these kind of problems we have used some undetermined multipliers introduced into our expression. Now we will treat these as independent and d n i's cannot be 0 therefore we have to set this equal to 0. That means now our next job is we write del log w del n i plus alpha minus beta e i equal to 0 we have to work on this now. Let us mark this we will be using this expression a bit later remember log w we have already got as n log n minus summation i n i log n i we have already derived this log w is equal to n log n minus summation n i log n i and if we were to find out its derivative then we need to find out the derivative of the first part and then the derivative of the second part with respect to d n i. Please keep this in mind that now we want to have the derivative of log w with respect to d n i. So, therefore, there are two parts one is derivative of n log n second is minus summation i n i log n i derivative of this. So, we will do this in two steps. Let us first work on the first one let us take the derivative of n log n with respect to n i and see what happens. So, part one del n log n this is what we want to find out what is this now it will be n pi n into derivative of n with respect to n i plus log n into derivative of n with respect to n i. Can I write like now 1 plus log n into del n by del n i. So, we need to worry about this n is a constant you know the thing can come in the mind that n total n is constant. So, a constant derivative of a constant should be equal to 0 therefore, this should be equal to 0 this is something which immediately comes to mind, but let us not act fast let us critically analyze. So, what we had is n is equal to n 0 plus n 1 plus n 2 plus so on you have n i plus so on. So, when I take del n by del n i remember that this n 0 n 1 n 2 they do not depend upon each other they are independent of each other. So, therefore, the derivative of the first term is going to be 0 second term is going to be 0 third is going to be 0 unless you reach n i where del n i by del n i is going to be equal to 1. So, that means this is equal to 1. So, we were working on you know del n log n by del n i and what we had done is n by n into del n del n i plus log n del n del n i which is equal to 1 plus log n del n del n i which we just find that this is equal to 1 therefore, this is equal to 1 plus log n. So, del n log n del n i is equal to 1 plus log n fine. So, first part of that is done now we need to work on the second part and what was that second part let us recall log w was equal to n log n minus summation i n i log n we have worked on the derivative of this. Now, let us work on the derivative of this. So, I am interested in now del del n i I want to take the derivative of this of summation instead of n i let me use j why I am using j is so that we do not confuse j with i we are taking derivative with respect to i. So, therefore, let me use j as a general term. So, this is equal to what now this is equal to summation j del del n i of n j log n j which is equal to summation j n j over n j del n j del n i plus log n j del n j over del n i which is equal to summation j in place y have 1 plus log n j into del n j by del n i. Now, when we need to expand when we expand let us say j starts from 0 1 2 etcetera remember that the number of molecules in let us say i plus 1 is not dependent upon number of molecules in the ground state etcetera they are independently in the different energy states. So, therefore, all the other terms when j is equal to 0 1 2 3 etcetera will disappear till j becomes i only when j is equal to i then del n i by del n i is equal to 1. So, therefore, the term which will survive is equal to 1 plus log n i n i not j n i when j is equal to i. So, remember that this is equal to 1 plus log n i. So, we have the expressions for both first part is 1 plus log n second part is 1 plus log n i and now we can go back and put into these this is what our you know aim was that we take the derivative of this and then work on that forward all right. So, let me work out here only our aim was del log w over del n i plus alpha minus beta e i equal to 0 and if we just go back del log w over del n i is this minus this right. If we take the derivative of this the derivative of the first part is 1 plus log n the derivative of the second part is equal to 1 plus log n i. So, we are going to use that and let us set over here 1 plus log n minus 1 minus log n i plus alpha minus beta e i equal to 0. This is equal to minus log n i upon n right plus alpha minus beta e i equal to 0. In other words log n i upon n is equal to alpha minus beta e i. Which I can rewrite as n i upon n is equal to exponential alpha minus beta e i we have n i upon n n i upon n is nothing, but population. But then if you look at the right hand side we have alpha we have beta these are the undetermined multipliers and we need to find out the expression for alpha and expression for beta. We can quickly find out the expression for alpha let us see how do we proceed. Let me use this space summation j n j I will use from here instead of i I will use j this is equal to I am taking summation summation n exponential alpha into exponential minus beta e j I am writing in terms of j ok n i is equal to n into exponential alpha into exponential minus beta instead of i I am writing j because I do not want to confuse i with the j. This one is equal to n total number of particles is equal to n it comes out exponential alpha constant comes out summation we have j exponential minus beta e j. So, therefore what we have exponential alpha is equal to 1 over summation j exponential minus beta e j. And once you substitute this into this form right remember that I can rewrite this as n i upon n is equal to exponential alpha into exponential minus beta e i. You substitute exponential alpha over here and what you get is this expression n i upon n is equal to exponential minus beta e i divided by summation j exponential minus beta e j this is Boltzmann distribution. What this equation tells is the population of an i th state n i is the number of molecules or number of particles in the i th state and n is the total number of particles. Therefore n i upon n is the population of i th state which can be calculated from the knowledge of energy levels or energy states and this summation. It is a very important result which is called Boltzmann distribution and this will form the basis for our future discussion when I introduce the concept of partition function. It is the partition function which is going to be connected with all the thermodynamic quantities and that will be the focus of discussion in the next lecture. Thank you very much.