 We now explore a qualitative understanding of the origin of this random force or random acceleration At from a gas kinetic model perspective. The Langevin equation that we wrote down in this reduced form as dv by dt equal to minus beta v plus a random acceleration At raises a question. The question is this, if you see the colloidal particle it experiences essentially molecular collisions. Both the systematic part or the viscous drag part as well as the random fluctuations coming from At originate from collisions itself. Once we have separately introduced therefore, a term At, why do we need the viscous drag term? Because random fluctuations should also include every other thing that happens because of collisions. The to understand this it is better to go through a simplified but illustrative model. In a nutshell we can say that the At term the random acceleration that accounts for the random force imparted on to the particle. However, the particle will also impart reaction to the fluid and the viscous resistance term comes because of that reaction back to the fluid. Qualitatively that is why these two have to be separated. However, if you look at the derivation of the viscous term it did not involve any molecular nature of matter. It came from Navier-Stokes equation their phenomenal phenomenological equation. You do not basically involve in their derivation any information on the either the temperature or the gas kinetic motion of the underlying molecules. Hence, that is also become a little intriguing for us to see how one can abruptly add a random term At in which part of the model has come from a continuum mechanics. Perhaps a better way of understanding the origin of both these terms is a straight away a gas kinetic model rather than a dense fluid like water model. So, in the gas kinetic model we can visualize fluctuations and gases also have viscous drags and hence both of them are related by kinetic theory easily and that therefore, should naturally lead us to the resolution of the total force experienced into systematic and random parts. So, we picture a situation now like this this is a gas kinetic model we have a particle. So, the particle is let us say massive much heavier than the gas molecule. So, its mass we call it as m p and let us say there is a along the x axis there is another gas molecule one molecule now which has a mass let us say m g g for gas is much smaller, but we assume its mass to be m g and the particle as m p and let us say that along the x direction the gas molecule let us say has a velocity v g and the particle has a velocity v p and the velocities are different. So, they collide and after collision this is before collision. So, after collision let us say the particle has a velocity v p prime and the gas molecules will have a velocity v g prime. So, 1 d collision problem point particles primes denote velocities after collision. So, the two body collisions are well known we can solve the equations to find out the velocities of both the particles after collision by employing momentum and energy conservation equations. So, the momentum conservation equation momentum has to be conserved along each direction the momentum conservation equation is m p mass of the particle and velocity of the particle that is the momentum of the particle before collision and m g v g the momentum of the gas molecule before collision should be equal to the momentum of the system after collision that is m p v p prime plus m g v g prime. So, this is the momentum conservation then we have kinetic energy conservation for elastic collisions. Let us assume elastic collisions. So, the kinetic energy is half of half will be common. So, half of m p v p square plus half m g v g square should be half m p v p prime square plus half m g v g prime square. So, this is the kinetic energy conservation. So, we have two equations we call the first one as equation 1 and the second one as equation 2. Our aim is to solve for the post collision values of the velocities of both the particle as well as the gas molecule. It can be done here because it is actually two it is a simultaneous equation with the two variables little complicated because one of them is non-linear quadratic type, but still it is easily solvable. We can substitute for example, for v g prime from the first one substitute we will get an equation and you can solve it. We will not go into the details of the solution, but the answers turn out to be the velocity of the particle post collision has the expression v p into m p minus m g plus 2 m g v g divided by total mass of the system m p plus m g. This is the post collision velocity of particle here. Similarly, v g prime will be v p will be replaced with v g and it will be m g minus m p here. This term will be 2 m p v p. Denominating it will be the same m g plus m p. This is the post collision velocity of gas molecule. We are not really interested in the absolute velocities of the gas molecule or the particle after collision. What we are keen is to know how much is the momentum change that has occurred in the particle due to one collision. So, quantity of interest change of momentum of the particle after one collision with the gas molecule. So, that is let us call it as delta p change in momentum of particle. This delta p by definition it is going to be mass of the particle into its momentum afterwards minus its momentum before the collision. So, it is m p v p prime minus m p v p. And from the expressions we just derived for m p v p prime here we can easily substitute them and see that this expression comes to 2 m p m g divided by m p plus m g into v g minus v p. So, it involves the masses of the particles and the velocity difference between the gas and the particle. Let us now simplify this special case of massive particle approximation apply massive particle approximation that is m p is much larger than m g. So, what is the physical basis for this? Simple a Brownian particle typically as we said 100 nanometers will have millions of molecules in it whereas, a gas molecule is just one molecule and hence we expect the particles to be millions times heavier even a factor of 100 is already a large number. So, as a result this assumption is perfectly justified and in such a limit we can see that this expression here m p m g by m p plus m g if m p is much larger in the denominator it is only m p which cancels with the m p here this m g in the denominator is neglected. So, we have the expression for the change in the momentum go to the next page. So, the change in the momentum of the particle is therefore, going to be 2 m g into v g minus v p. This is the expression for the momentum imparted to the particle or we can say change in momentum due to one collision this is particle. Now, we obtained this expression within a one dimensional framework. So, in a one dimensional framework my particle it is experiencing collision either in this direction or from this direction. My v g can be hitting the front side of the particle or it can be hitting from the back side. All that it says is the relative velocity between them is what causes a momentum change. If strictly both of them are having the same velocity then there will be no momentum imparted no momentum change occurring because they will not collide at all it is the relative velocity that makes the difference. This is a very important a very simple, but very important information that we obtained. Now, let us go a step further since we have decided to use the notation v for velocity and for future simplification first of all we simplify notation just for purpose of simplification the gas velocity the v g we denote it just by g. And particle velocity v p we denote it by just v because we decided to use v in the Stokes law. So, we standardize our notation. And let us now say that an ith collision has occurred one particular collision which means my velocity is going to be v g i or we decided to call it as g i. This collision could have occurred either from the front of the particle that is along the direction of v p or it could have occurred on the lee side of the particle since gas molecules have random velocities when we resolve the consider only the x directional components it can be either from the front or from the back of the particle and it has therefore a random nature. In order to obtain an average force experienced or average momentum transferred per collision one has to average over all these realizations of possible realizations of particle a molecular motion a molecular collision. So, if an ith collision has occurred for example, so we can write the above expression more specifically delta p i equal to 2 mg into g i minus v in the new notation g i is the molecular velocity and v is the particle velocity. To obtain an average behavior for example, we have to ascribe the statistical properties to g i g i here random velocity of molecule. Let us say that basically it comes from a Maxwell's distribution. So, it has a Gaussian nature along the x direction if we see it will have a certain value g i let us say to be simple let us say it has a velocity plus c with the probability half where c is an average gas kinetic speed or it has a minus c with the gain probability half. So, g i is a at the moment we are keeping it as a dichotomous quantity, but the important thing is there are as many forward collisions or as backward collisions. So, even if we consider them as a distributed quantity this property has to be preserved. So, which basically means if I do ensemble average or for all average overall ensembles then the average velocity of the molecule along the direction should be 0. Why? Because the fluid is stationary since gas is stationary macroscopically or on the bulk. If g i is not 0 if it had a certain component along the x direction the gas would have been seen as moving along that direction which is not the case. And the second important property is that the velocity imparted in one collision is not correlated with the velocity imparted in another collision that is we can say that g i g j will be 0 mutually uncorrelated. If we use this property then the ensemble average momentum transfer or momentum changed is delta p i average which is going to be 2 mg into average of g i minus 2 mg average of v. And v does not depend on molecule fluctuations because that is the average velocity that we imparted to the particle and therefore, it is going to be 0 minus 2 mg into v. So, this is a very important result. So, the average momentum transferred is equal to minus 2 mg into which is in the opposite direction of the velocity of the particle. If the particle is moving in the forward direction then the momentum change is negative the velocity is moving in the backward direction the change in velocities positive. So, it is basically in the direction opposite to the velocity. This now throws us a hint on the origin of viscous drag without using Navier-Stokes equation. We have considered average momentum change per collision. Now, let us say that the gas there are n g molecules per unit volume semi ta cube and let us say nu g the rate of collision or the frequency of collision the particles or the number of collisions number of collisions per unit time say per second on the particle. We showed that one collision leads to a change of delta p i average and therefore, the momentum rate of change of momentum of particle that should be equal to the number of molecules the striking per unit time and the momentum change the per strike per particle per molecule. So, that is delta p average which comes out to be minus of 2 mg nu g into v. The rate of change of momentum we identify as the force experienced by the particle. Therefore, the average force experienced by the particle equal to F drag force is minus of mg nu g into v. This is law of drag force from gas kinetic perspective without using Navier-Stokes. Obviously, we do not the expression to correspond to that based on Navier-Stokes equation, but the most important thing to learn is from this perspective as well one obtains a systematic force coming due to molecular collisions which is proportional to the velocity of the particle and directed in the opposite direction. Just to complete this expression we can identify the kinetic friction coefficient friction coefficient F 1 as 2 mg nu g. Therefore, one can say that the drag force is minus of F 1 into v. We have deliberately used subscript 1 here for the friction coefficient because F 1 need not be same as F that is coming from Navier-Stokes perspective. In other words, the drag force that a particle experiences in a dense fluid need not have need not be numerically same as that it experiences in a dilute gas. The functional form itself could be different. This is all pointing out to that. Our aim is not to derive an expression for the drag force, but to point out that molecular collisions naturally lead us to a force systematic force term upon ensemble averaging and that force is proportional to velocity. The F 1 incidentally that we get if we substitute the various quantities we can say that it is a 2 mg and then there are collisions which are taking place is nothing, but the gas density into the velocity of the particle. And if I use this into of course, the cross sectional area subtended by the sphere which is pi a square since we are talking of only x direction. So, we can actually show that all these will come to rho g into particle diameter square into square root of 2 pi kT by m. All that we have done is to use the expression that C equal to root 8 kT by pi m is the mean thermal speed and of course, we have used the definition of bulk density rho g as the number density into the mass of the molecule. So, with this we can obtain this expression to point out considering that we did only 1D specular reflection normal incidence analysis. This expression is very close to the exact expression for a gas kinetic formula is F 1 equal to 2 third rho g T square into 2 pi kT by m. This is the Epstein formula. So, we are not too bad we are very close a factor of 2 third considering the simplicity of our model is quite agreeable. Main point is we get all the correct dependencies with respect to particle size density etc. So, now that we derived a systematic part from gas kinetic considerations we will move over to how to construct a random acceleration from the same model we do it in our next lecture. Thank you.