 Welcome to module 6 of chemical kinetics and transition state theory. Today we are going to look at the distribution of speed and velocities. In last module we had looked at the Boltzmann distribution itself, but what can we do with that Boltzmann distribution is what we are going to look at today. In doing so, we will also cover up one important topic that will be important later on which is the translational partition function. So, a quick recap of module 5, we had given you a partial derivation of calculating the equilibrium density matrix. It is given by this Boltzmann distribution e to the power of minus beta h where h is the Hamiltonian divided by q and q is called the partition function which is an integral over all space dp dq into e to the power of minus beta h. And we also discussed how to be calculate average of any property. If I have any quantity let us say a as a function of some q and p in general to find the average of a, I must integrate over all p and q rho which gives me the probability of being at q and p multiplied by a which gives me the value at q and p. So, in the last module we looked at a few examples which is average momentum and average kinetic energy. So, before moving forward let us do one thing which is going to be useful. So, I have written down the definition of our rho equilibrium which is what we derived in the last module e to the power of minus beta h divided by what this integral is basically the q, written the Hamiltonian explicitly which is the kinetic energy plus the potential energy. So, I want to just substitute this h in this rho equilibrium and see if we can simplify this equation a little bit. So, if I substitute what I get is. So, I write h as sum over i p i square over 2 m plus v of q divided by integral of dq integral of dp e to the power of minus beta I write the same big summation here and summation in exponential can be converted into a product of exponentials. So, I write the same thing as e to the power of minus beta sum over i p i square over 2 m into e to the power of minus beta v of q divided by and I will do the same separation in the denominator. I will write the momentum first because I have momentum first in the numerator. So, just to be consistent we write it as this my bad it should be an integral not a summation. So, let us remove the summation let us put in the integral dq e to the power of minus beta v of q. So, what you notice that we have separated the terms as a function of p and as a function of q. So, this part we will define to be rho equilibrium of p only p no there is no q dependence in this. This portion I will define of q there is no p dependence in this one. So, in total I am writing this rho equilibrium of q comma p in separable form of rho equilibrium of q into rho equilibrium of p where rho equilibrium of p is defined here and rho equilibrium of q is defined here. So, let us look at the equilibrium function for p for a moment we can actually simplify it a little bit more. So, I have rewritten the rho equilibrium of p right here. Now, the denominator that we have here my bad this should be some over i p i square over 2 m this should be some over i p i square over 2 m. So, this denominator that we get is called the translational partition function why translational because it is simply kinetic energy. So, this is the partition function when v of q is 0 for a free particle ok. So, this partition function is called the translational partition function and we can actually simplify it and calculate it we can find an closed form answer for this. So, let us try to do that. So, this thing first of all note that integral over dp vector is the same thing as integral over dp 1 integral over dp 2 till till integral over dp 3 m remember we have 3 n momenta 3 n positions why 3 n n is the number of particles for each particle I have x y and z. So, I have p x p y p z for particle 1 p x p y p z for particle 2 there on so forth. So, I have 3 n momenta. So, this is what I have that I have to calculate well we note that I can simplify this integral as e to the power of minus beta p 1 square over 2 m into e to the power of minus beta p 2 square over 2 m till e to the power of minus beta p 3 n square over 2 m. So, I have again taken the exponential exponential was a sum form. So, I can take it into a product form ok. Now, I note that this is equal to so, I separate all the terms out take them to the corresponding integral. So, let us look at each integral separately. So, I separated out all the integrals all are independent integrals and I can evaluate them one at a time that is the beauty of this. And this one integral is called a Gaussian integral from all overall space minus infinity to plus infinity that integral form is known I have provided you the formula and again throughout the course any integral which is complex we will provide you you do not have to memorize any of this this is not really a max course ok. So, this thing here we note that in my formula I have provided you this a. So, if a here will be beta over 2 m yeah it is the constant before my variable which is nothing but 1 over 2 m k B. So, this thing becomes equal to root of 2 pi m k B. So, I am simply using this formula here root of pi over a. So, root of pi divided by a, but a note is 1 over something. So, I get this thing, but I get the same integral when I solve for P2. In fact, I will get the same for all 3 n of them. So, if I multiply these together what I get is or I just write slightly differently. So, now we have derived the formula for the translational partition function this will be useful later on. So, we will keep it for now and once we use it I will remind you of this. For now this is nothing but the denominator of rho equilibrium of P. So, now finally, I get rho equilibrium of P is equal to e to the power of minus beta sum over i P i square over 2 m over 2 pi k B T m to the power of 3 n over 2. So, let us just look at it in one dimension only. So, for this I get rho equilibrium. So, for n equal to 1 I still have 3 momentum p x p y p z and in our notation we refer p 1 as p x p 2 as p y and p 3 as p z and here I will substitute n equal to 1. So, I get this. So, that is your distribution of momentum in one dimension. Trying to figure out what will be the distribution of speed. After I do not care what is the value of momentum in different directions x y and z speed is a more natural quantity. I want to figure out how fast is the particle moving yeah that is a very natural question to ask. So, that is what I am trying to figure out. So, let us get to that. First of all let us be formal and define speed. Speed is nothing but root of v 1 square plus v 2 square plus v 3 square where v 1 v 2 v 3 are the velocities in x y and z direction. But I write it in the language of momentum and I call this as modulus of p divided by n where modulus of p. So, I am trying to find a distribution of u and how do I do that. All right. So, let us just look at rho once more of p 1 p 2 p 3. This is equal to e to the power of minus beta just from last slide. The important thing to remember when converting between distributions. So, I am going from a distribution of p 1 p 2 p 3 to a distribution of u. When you do that in whichever field not only kinetics or thermodynamics, quantum dynamics wherever, remember the volume element that is very important. So, I will start writing that a little bit more explicitly. So, this is really what this probability density anyway mean. In a very small volume of size dp 1 dp 2 dp 3 what is the probability of finding the system there. So, I am asking you the question what is the probability rho u du is equal to. We will actually start with slightly different question which is what is mod p dp. Because here I have everything in the language of momentum. So, I will start with momentum and then I will go back and answer this question. Now, to calculate this I will have to remind you a little bit of what is called the spherical polar coordinates. You must have seen this in some form or another. It occurs in many many different contexts. So, imagine you have three dimensions x, y and z. I define a new coordinate system which is called r, theta and phi. R is nothing but what we are looking for in the previous. Theta is the angle that the vector r makes with the z axis and phi is the angle that the projection of the vector r on the xy plane makes with the x axis. So, a few important properties that I want to remind you of r goes from 0 to infinity, theta goes from 0 to pi and phi goes from 0 to 2 pi. These are the limits and more importantly what is the volume element in this. So, dx dy dz is equal to r square sin theta dr d theta d phi. So, now the rho equilibrium I had already written in the last slide. I have forgotten to write dp1, dp2, dp3. I want to find rho equilibrium p theta phi dp d theta d phi. That is my first challenge. So, that you can see e to the power of minus beta, this you note is mod p square over 2m. Once more, mod p is nothing but root of p1 square plus p2 square plus p3 square divided by 2 pi kb tm to the power of 3 half. And now we introduce the volume element that we talked of in the last slide which is mod of p square sin theta dp d theta d phi. But this is not what we wanted. We wanted rho equilibrium of just p. I do not care which angle theta and phi it is. Anyway, theta and phi are arbitrary. They depend on my axis choice. So, what I am going to do is I integrate over all of theta and all of phi. So, because I want the average value of theta and phi. Again, the range of theta is 0 to pi, phi is 0 to 2 pi. And I write this whole thing here and I write dp or I should have a dp here as well. So, what I do is I take the terms that are independent of theta and phi outside the integral. And then I integrate over theta and phi. So, for theta you note I have sin theta for phi I have d phi. I leave it as a homework for you to prove this is equal to 2 and this is equal to 2 pi. These are very easy integrals you should be able to do. So, in total I get 4 pi mod rho square mod p square I am sorry 2 pi kbtm to the power of 3 half e to the power of minus beta p square over. So, I calculate this rho equilibrium I have just copied it from the last slide, but I wanted rho of u not p. So, well that is not very hard now. I note that u is defined to be mod of p over m and so, du is d mod of p over m. And so, we just substitute these quantities here p is mu dp is m du. So, this becomes equal to e to the power of minus beta 1 over 2 m mod p square which is m square u square 2 pi kbtm to the power of 3 half 4 pi p square which is nothing but m square u square dp which is nothing but m du. This is rho equilibrium of u du. So, I just simplify this a little bit this m cancels with this just massaging a little bit here and there I get 4 pi u square. And I just transform this into this form. You can quickly verify whether I have written it correctly or not. Make sure you can do this if I have made a mistake then please correct. So, this is the final form I get for the Maxwell Boltzmann distribution. It is a very famous distribution it is named after these two outstanding scientists Maxwell and Boltzmann who have contributed immensely towards statistical mechanics. So, let us just look at one thing let us make a plot of it because I just want to look at as a function of u how does rho equilibrium of u look like. So, well if you look at it this function is a product of two different functions one is u square. So, u square looks like this. So, I am just qualitatively trying to find how the curve will look like and the other function is this Gaussian and this Gaussian looks like this. This will keep on going till it reaches 0 at infinity very qualitatively I am not being very precise here. So, if I take a product of these two my question is what will you get. So, I would recommend pause the video take a moment and multiply this on yourself do not take help of any computer or anything and make a plot. Hopefully you pause the video and made the plot by your own self. So, the point to note is that rho equilibrium I should change the color rho equilibrium of u equal to 0 is 0 if I put u equal to 0 this red thing goes to 0 and rho equilibrium of u equal to infinite is also 0 because the green thing goes to 0. So, I start with 0 here and I must end with 0 as well. So, I get a function that increases initially and then over time it decreases and goes to 0. So, this is the Boltzmann Maxwell Boltzmann distribution. So, I have a challenge to you a question you have this distribution good in the previous module we calculated the average momentum and we showed it is equal to 0. What is your guess what will be the average speed will it be 0 greater than 0 or less than 0. So, please go to this link that is provided here and answer there this is completely anonymous this is for your own good and you will get an immediate feedback based on your answer. So, hopefully all of you have answered this question based on whatever you think it is right let us try to solve it now. So, the average u well how do I find average u it is a same trick integral over du rho equilibrium of u into u. Now, the question is what is the limit for u is it minus infinity to infinity or 0 to infinity it is 0 to infinity remember that u is magnitude it can never be negative we are only dealing with the overall quantity. And so, we substitute all of this big formula here 4 pi u square e to the power of minus beta m u square over 2 into u. So, we take all the constants out of the integral and what we are left with is 0 to infinity du u cube e to the power of minus beta m over 2 u square. And again as before we will provide you the integrals when you need the integrals this is slightly complex integral and so, we have provided you the answer here on how to solve. So, to match these 2 my a here will be beta m over 2. So, I get m over 2 pi k t 3 half into 4 pi and 1 over 2 a square 2 a is I will write this as m over 2 k b t because I have k b t in this equation. So, m square into k b t. So, you can go ahead and simplify this equation and show this is equal to root of 8 k b t over pi. So, this is clearly greater than 0. So, by the way you did not had to do the match to tell whether it is greater than 0 or not. See speed is a positive quantity. So, if I am averaging over a lot of particles over positive numbers only well you are going to get some positive number right. So, it cannot be 0 it cannot be less than 0 I have some speed of some particle which is positive and so, you can formally show it is equal to this. So, in summary today we have looked at the Maxwell Boltzmann distribution of speed. It is a very important distribution and this will be very useful in the coming modules when we discuss kinetic theory of collisions. We have shown that the average speed is equal to root of k 8 k t over pi m. Please do not memorize any of these as and when needed we will always be providing you equations. This is not a memory going to be a memory test this is not going to be a mathematics test. Finally, one another thing just keep in mind for future we have also derived what is called a translational partition function as m over 2 pi k t to the power of 3 over 3 and over 2. Thank you very much.